Thinking Process

If $\boldsymbol{{}\alpha}$ is the one of the zeroes of the quadratic polynomial $f(x)=a x^{2}+b x+c$. Then, $f(\boldsymbol{{}\alpha})$ must be equal to 0 .

Solution

(a) Given that, one of the zeroes of the quadratic polynomial say $p(x)=(k-1) x^{2}+k x+1$ is -3 , then

$ p(-3)=0 $

$\Rightarrow$ $ (k-1)(-3)^{2}+k(-3)+1=0 $

$\Rightarrow$ $ 9(k-1)-3 k+1=0 $

$\Rightarrow$ $ 9 k-9-3 k+1=0 $

$\Rightarrow$ $ 6 k-8=0 $

$ \therefore \quad k=4 / 3 $

Solution

(c) Let $a x^{2}+b x+c$ be a required polynomial whose zeroes are -3 and 4 .

Then, sum of zeroes $=-3+4=1$

$\Rightarrow \quad \dfrac{-b}{a}=\dfrac{1}{1} \Rightarrow \dfrac{-b}{a}=-\dfrac{(-1)}{1}$

$\because$ sum of zeroes $=\dfrac{-b}{a}$

and

product of zeroes $=-3 \times 4=-12$

$\because$ product of zeroes $=\dfrac{c}{a}$

$\Rightarrow \quad \dfrac{c}{a}=\dfrac{-12}{1}$

From Eqs. (i) and (ii),

$ \begin{aligned} a & =1, b=-1 \text{ and } c=-12 \\ & =a x^{2}+b x+c \end{aligned} $

$\therefore$ Required polynomial $=1 \cdot x^{2}-1 \cdot x-12$

$ \begin{aligned} & =x^{2}-x-12 \\ & =\dfrac{x^{2}}{2}-\dfrac{x}{2}-6 \end{aligned} $

We know that, if we multiply/divide any polynomial by any constant, then the zeroes of polynomial do not change.

Alternate Method

Let the zeroes of a quadratic polynomial are $\alpha=-3$ and $\beta=4$.

Then, sum of zeroes $\quad=\alpha+\beta=-3+4=1$

and product of zeroes $=\alpha \beta=(-3)(4)=-12$

$\therefore \quad$ Required polynomial $=x^{2}-$ (sum of zeroes) $x+$ (product of zeroes)

$ \begin{aligned} & =x^{2}-(1) x+(-12)=x^{2}-x-12 \\ & =\dfrac{x^{2}}{2}-\dfrac{x}{2}-6 \end{aligned} $

Solution

(d) Let $p(x)=x^{2}+(a+1) x+b$

Given that, 2 and -3 are the zeroes of the quadratic polynomial $p(x)$.

$$ \begin{matrix} \therefore & p(2)=0 \text{ and } p(-3)=0 \\ \Rightarrow & 2^{2}+(a+1)(2)+b=0 \\ \Rightarrow & 4+2 a+2+b=0 \\ \Rightarrow & 2 a+b=-6 \\ \text{ and } & (-3)^{2}+(a+1)(-3)+b=0 \\ \Rightarrow & 9-3 a-3+b=0 \\ \Rightarrow & 3 a-b=6 \tag{ii} \end{matrix} $$

On adding Eqs. (i) and (ii), we get

$ 5 a=0 \Rightarrow a=0 $

Put the value of $a$ in Eq. (i), we get

$ 2 \times 0+b=-6 \Rightarrow b=-6 $

So, the required values are $a=0$ and $b=-6$.

Solution

(d) Let $p(x)=a x^{2}+b x+c$ be the required polynomial whose zeroes are -2 and 5 .

$$ \begin{align*} \therefore \quad \text{ Sum of zeroes } & =\dfrac{-b}{a} \\ \Rightarrow \quad \dfrac{-b}{a} & =-2+5=\dfrac{3}{1}=\dfrac{-(-3)}{1} \tag{i} \end{align*} $$

and $\quad$ product of zeroes $=\dfrac{c}{a}$

$\Rightarrow \quad \dfrac{c}{a}=-2 \times 5=\dfrac{-10}{1}$

From Eqs. (i) and (ii),

$ \begin{aligned} a=1, b & =-3 \text{ and } c=-10 \\ \therefore \quad p(x) & =a x^{2}+b x+c=1 \cdot x^{2}-3 x-10 \\ & =x^{2}-3 x-10 \end{aligned} $

But we know that, if we multiply/divide any polynomial by any arbitrary constant. Then, the zeroes of polynomial never change.

$ \begin{matrix} \therefore & p(x)=k x^{2}-3 k x-10 k \quad \text{ [where, } k \text{ is a real number] } \\ \Rightarrow & p(x)=\dfrac{x^{2}}{k}-\dfrac{3}{k} x-\dfrac{10}{k}, \quad \text{ [where, } k \text{ is a non-zero real number] } \end{matrix} $

Hence, the required number of polynomials are infinite i.e., more than 3.

Thinking Process

Firstly, we find the sum of product of two zeroes at a time and put the value of one of the zeroes i.e.,zero, we get the required product of the other two zeroes.

Solution

(b) Let $p(x)=a x^{3}+b x^{2}+c x+d$

Given that, one of the zeroes of the cubic polynomial $p(x)$ is zero.

Let $\alpha, \beta$ and $\gamma$ are the zeroes of cubic polynomial $p(x)$, where $a=0$.

We know that,

Sum of product of two zeroes at a time $=\dfrac{C}{a}$

$\Rightarrow \quad \alpha \beta+\beta \gamma+\gamma \alpha=\dfrac{c}{a}$

$\Rightarrow \quad 0 \times \beta+\beta \gamma+\gamma \times 0=\dfrac{c}{a} \quad[\because \alpha=0$, given $]$

$\Rightarrow \quad 0+\beta \gamma+0=\dfrac{c}{a}$

$\Rightarrow \quad \beta \gamma=\dfrac{c}{a}$

Hence, product of other two zeroes $=\dfrac{c}{a}$

Thinking Process

Firstly, we find the value of constant term ’ $c$ ‘, by using $p(-1)=0$. After that we find the product of all zeroes and put the value of one of the zeroes. Finally, we get the required result.

Solution

(a) Let $p(x)=x^{3}+a x^{2}+b x+c$

Let $\alpha, \beta$ and $\gamma$ be the zeroes of the given cubic polynomial $p(x)$.

[given]

$ \begin{matrix} \Rightarrow & (-1)^{3}+a(-1)^{2}+b(-1)+c & =0 \\ \Rightarrow & -1+a-b+c & =0 \\ \Rightarrow & & c & =1-a+b \end{matrix} $

We know that,

Product of all zeroes $=(-1)^{3} \cdot \dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{3}}=-\dfrac{C}{1}$

$ \alpha \beta \gamma=-c $

$ \Rightarrow \quad(-1) \beta \gamma=-c \quad[\because \alpha=-1] $

$$ \begin{matrix} \Rightarrow & \beta \gamma=c \\ \Rightarrow & \beta \gamma=1-a+b \tag{i} \end{matrix} $$

Hence, product of the other two roots is $1-a+b$.

Alternate Method

Since, -1 is one of the zeroes of the cubic polynomial $f(x)=x^{2}+a x^{2}+b x+c$ i.e., $(x+1)$ is a factor of $f(x)$.

Now, using division algorithm,

$ \begin{aligned} & \dfrac{x^{2}+(a-1) x+(b-a+1)}{\dfrac{x^{3}+a x^{2}+b x+c}{(a-1) x^{2}+b x}} \\ & \dfrac{(a-1) x^{2}+(a-1) x}{(b-a+1) x+c} \\ & \dfrac{(b-a+1) x(b-a+1)}{(c-b+a-1)} \end{aligned} $

$\therefore x^{3}+a x^{2}+b x+c=(x+1) \times{x^{2}+(a-1) x+(b-a+1)}+(c-b+a-1)$

$\Rightarrow x^{3}+a x^{2}+b x+(b-a+1)=(x+1){x^{2}+(a-1) x+(b-a+1)}$

Let $\alpha$ and $\beta$ be the other two zeroes of the given polynomial, then

Product of zeroes $=(-1) \alpha \cdot \beta=\dfrac{\text{ - Constant term }}{\text{ Coefficient of } x^{3}}$

$ \begin{aligned} \Rightarrow & -\alpha \cdot \beta & =\dfrac{-(b-a+1)}{1} \\ \Rightarrow & \alpha \beta & =-a+b+1 \end{aligned} $

Hence, the required product of other two roots is $(-a+b+1)$.

Solution

(b) Let given quadratic polynomial be $p(x)=x^{2}+99 x+127$.

On comparing $p(x)$ with $a x^{2}+b x+c$, we get

$ a=1, b=99 \text{ and } c=127 $

We know that,

$ \begin{aligned} x & =\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ & =\dfrac{-99 \pm \sqrt{(99)^{2}-4 \times 1 \times 127}}{2 \times 1} \\ & =\dfrac{-99 \pm \sqrt{9801-508}}{2} \\ & =\dfrac{-99 \pm \sqrt{9293}}{2}=\dfrac{-99 \pm 96.4}{2} \\ & =\dfrac{-99+96.4}{2}, \dfrac{-99-96.4}{2} \\ & =\dfrac{-2.6}{2}, \dfrac{-195.4}{2} \\ & =-1.3,-97.7 \end{aligned} $

[by quadratic formula]

Hence, both zeroes of the given quadratic polynomial $p(x)$ are negative.

Alternate Method

We know that,

In quadratic polynomial, if $\begin{gathered}a>0 \\ a<0\end{gathered}$ or $\begin{aligned} & b>0, c>0 \\ & b<0, c<0\end{aligned}$, then both zeroes are negative.

In given polynomial, we see that

$ a=1>0, b=99>0 \text{ and } c=127>0 $

which satisfy the above condition.

So, both zeroes of the given quadratic polynomial are negative.

Solution

Let $$ p(x)=x^2+k x+k, k \neq 0 $$

On comparing $p(x)$ with $a x^2+b x+c$, we get $$ \begin{aligned} a & =1, b=k \text { and } c=k \\ x & =\dfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\ & =\dfrac{-k \pm \sqrt{k^2-4 k}}{2 \times 1} \\ & =\dfrac{-k \pm \sqrt{k(k-4)}}{2}, k \neq 0 \end{aligned} $$ [by quadratic formula]

Here, we see that $$ \begin{aligned} & k(k-4)>0 \\ & \Rightarrow \quad k \in(-\infty, 0) \cup(4, \infty) \\ & \end{aligned} $$

Now, we know that In quadratic polynomial $a x^2+b x+c$ If $a>0, b>0, c>0$ or $a<0, b<0, c<0$, then the polynomial has always all negative zeroes. and if $a>0, c<0$ or $a<0, c>0$, then the polynomial has always zeroes of opposite sign.

$$ \begin{array}{cll} \text { Case I If } & k \in(-\infty, 0) \text { i.e., } k<0 \\ & \Rightarrow \quad a=1>0, \quad b, c=k<0 \\ & \end{array} $$

So, both zeroes are of opposite sign. $$ \begin{array}{cll} \text { Case II If } & k \in(4, \infty) \text { i.e., } k \geq 4 \\ \Rightarrow & a=1>0, b, c \geq 4 \end{array} $$

So, both zeroes are negative.

Hence, in any case zeroes of the given quadratic polynomial cannot both be positive.

Solution

(c) The zeroes of the given quadratic polynomial $a x^{2}+b x+c, c \neq 0$ are equal. If coefficient of $x^{2}$ and constant term have the same sign i.e., $c$ and a have the same sign. While $b$ i.e., coefficient of $x$ can be positive/negative but not zero.

$ \begin{matrix} \text{ e.g., } & \text{ (i) } x^{2}+4 x+4=0 & \text{ (ii) } & x^{2}-4 x+4=0 \\ \Rightarrow & (x+2)^{2}=0 & \Rightarrow & (x-2)^{2}=0 \\ \Rightarrow & x=-2,-2 & \Rightarrow & x=2,2 \end{matrix} $

Alternate Method

Given that, the zeroes of the quadratic polynomial $a x^{2}+b x+c$, where $c \neq 0$, are equal i.e., discriminant $(D)=0$

$ \begin{matrix} \Rightarrow & b^{2}-4 a c=0 \\ \Rightarrow & b^{2}=4 a c \\ \Rightarrow & a c=\dfrac{b^{2}}{4} \\ \Rightarrow & a c>0 \end{matrix} $

which is only possible when $a$ and $c$ have the same signs.

Solution

(a) Let

$ p(x)=x^{2}+a x+b . $

Now, $\quad$ product of zeroes $=\dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{2}}$

Let $\alpha$ and $\beta$ be the zeroes of $p(x)$.

$\therefore \quad$ Product of zeroes $(\alpha \cdot \beta)=\dfrac{b}{1}$

$$ \begin{equation*} \Rightarrow \quad \alpha \beta=b \tag{i} \end{equation*} $$

Given that, one of the zeroes of a quadratic polynomial $p(x)$ is negative of the other.

$ \begin{array}{ll} \therefore & \alpha \beta<0 \\ \text{ So, } & b<0 \end{array} $

[from Eq. (i)]

Hence, $b$ should be negative

Put $a=0$, then,

$\Rightarrow$

$\Rightarrow$ then it has no linear term i.e., $a=0$ and the constant term is negative i.e., $b<0$.

Alternate Method

Let

$ f(x)=x^{2}+a x+b $

and by given condition the zeroes are $\alpha$ and $-\alpha$.

$\therefore$ Sum of the zeroes $=\alpha-\alpha=a$

$\Rightarrow$

$$ \begin{align*} p(x) & =x^{2}+b=0 \\ x^{2} & =-b \\ x & = \pm \sqrt{-b} \end{align*} $$

$\therefore \quad f(x)=x^{2}+b$, which cannot be linear.

and product of zeroes $=\alpha \cdot(-\alpha)=b$

$\Rightarrow \quad-\alpha^{2}=b$

which is possible when, $b<0$.

Hence, it has no linear term and the constant term is negative.

Solution

(d) For any quadratic polynomial $a x^{2}+b x+c, a \neq 0$, the graph of the Corresponding equation $y=a x^{2}+b x+c$ has one of the two shapes either open upwards like $u$ or open downwards like $n$ depending on whether $a>0$ or $a<0$. These curves are called parabolas. So, option (d) cannot be possible.

Also, the curve of a quadratic polynomial crosses the $X$-axis on at most two points but in option (d) the curve crosses the $X$-axis on the three points, so it does not represent the quadratic polynomial.

Solution

(i) No, because whenever we divide a polynomial $x^{6}+2 x^{3}+x-1$ by a polynomial in $x$ of degree 5 , then we get quotient always as in linear form i.e., polynomial in $x$ of degree 1 . Let divisor $=$ a polynomial in $x$ of degree 5

$ \begin{aligned} & =a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f \\ \text{ quotient } & =x^{2}-1 \end{aligned} $

and

$ \text{ dividend }=x^{6}+2 x^{3}+x-1 $

By division algorithm for polynomials,

Dividend $=$ Divisor $\times$ Quotient + Remainder

$ \begin{aligned} & =(a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f) \times(x^{2}-1)+\text{ Remainder } \\ & =(\text{ a polynomial of degree } 7)+\text{ Remainder } \end{aligned} $

[in division algorithm, degree of divisor $>$ degree of remainder] $=($ a polynomial of degree 7$)$

But dividend $=$ a polynomial of degree 6

So, division algorithm is not satisfied.

Hence, $x^{2}-1$ is not a required quotient.

(ii) Given that, Divisor $p x^{3}+q x^{2}+r x+s, p \neq 0$

and dividend $=a x^{2}+b x+c$

We see that,

Degree of divisor $>$ Degree of dividend

So, by division algorithm,

quotient $=0$ and remainder $=a x^{2}+b x+c$

If degree of dividend $<$ degree of divisor, then quotient will be zero and remainder as same as dividend.

(iii) If division of a polynomial $p(x)$ by a polynomial $g(x)$, the quotient is zero, then relation between the degrees of $p(x)$ and $g(x)$ is degree of $p(x)<$ degree of $g(x)$.

(iv) If division of a non-zero polynomial $p(x)$ by a polynomial $g(x)$, the remainder is zero, then $g(x)$ is a factor of $p(x)$ and has degree less than or equal to the degree of $p(x)$. i.e., degree of $g(x) \leq$ degree of $p(x)$. (v) No, let $p(x)=x^{2}+k x+k$ If $p(x)$ has equal zeroes, then its discriminant should be zero.

$$ \begin{equation*} \therefore \quad D=B^{2}-4 A C=0 \tag{i} \end{equation*} $$

On comparing $p(x)$ with $A x^{2}+B x+C$, we get

$\begin{aligned} & A=1, B=k \text { and } C=k \\ & \therefore \quad(k)^2-4(1)(k)=0 \\ & \Rightarrow \quad k(k-4)=0 \\ & \Rightarrow \quad k=0,4 \\ & \end{aligned}$

So, the quadratic polynomial $p(x)$ have equal zeroes only at $k=0,4$.

Solution

(i) False, if the zeroes of a quadratic polynomial $a x^{2}+b x+c$ are both positive, then

$ \alpha+\beta=-\dfrac{b}{a} \quad \text{ and } \alpha \cdot \beta=\dfrac{c}{a} $

where $\alpha$ and $\beta$ are the zeroes of quadratic polynomial.

$ \begin{aligned} & \therefore \quad c<0, a<0 \text{ and } b>0 \\ & \text{ or } \quad c>0, a>0 \text{ and } b<0 \end{aligned} $

(ii) True, if the graph of a polynomial intersects the $X$-axis at only one point, then it cannot be a quadratic polynomial because a quadratic polynomial may touch the $X$-axis at exactly one point or intersects $X$-axis at exactly two points or do not touch the $X$-axis.

(iii) True, if the graph of a polynomial intersects the $X$-axis at exactly two points, then it may or may not be a quadratic polynomial. As, a polynomial of degree more than $z$ is possible which intersects the $X$-axis at exactly two points when it has two real roots and other imaginary roots. (iv) True, let $\alpha, \beta$ and $\gamma$ be the zeroes of the cubic polynomial and given that two of the zeroes have value 0 .

$ \begin{aligned} & \text{ Let } \quad \beta=\gamma=0 \\ & \text{ and } \\ & f(x)=(x-\alpha)(x-\beta)(x-\gamma) \\ & =(x-\alpha)(x-0)(x-0) \\ & =x^{3}-a x^{2} \end{aligned} $

which does not have linear and constant terms.

(v) True, if $f(x)=a x^{3}+b x^{2}+c x+d$. Then, for all negative roots, $a, b, c$ and $d$ must have same sign.

(vi) False, let $\alpha, \beta$ and $\gamma$ be the three zeroes of cubic polynomial $x^{3}+a x^{2}-b x+c$.

Then,

$ \text{ product of zeroes }=(-1)^{3} \dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{3}} $

$$ \begin{matrix} \Rightarrow & \alpha \beta \gamma=-\dfrac{(+c)}{1} \\ \Rightarrow & \alpha \beta \gamma=-c \tag{i} \end{matrix} $$

Given that, all three zeroes are positive. So, the product of all three zeroes is also positive i.e., $\alpha \beta \gamma>0$

$\Rightarrow \quad-c>0$

[from Eq. (i)]

$\Rightarrow$

$c<0$

Now, sum of the zeroes $=\alpha+\beta+\gamma=(-1) \dfrac{\text{ Coefficient of } x^{2}}{\text{ Coefficient of } x^{3}}$

$ \Rightarrow \quad \alpha+\beta+\gamma=-\dfrac{a}{1}=-a $

But $\alpha, \beta$ and $\gamma$ are all positive

Thus, its sum is also positive.

So,

$\alpha+\beta+\gamma>0$

$\Rightarrow$

$-a>0$

$\Rightarrow$

$a<0$

and sum of the product of two zeroes at a time $=(-1)^{2}$. Coefficient of $x=\dfrac{-b}{\text{ Coefficient of } x^{3}}$

$ \begin{matrix} \Rightarrow & \alpha \beta+\beta \gamma+\gamma \alpha=-b \\ \because & \alpha \beta+\beta \gamma+\alpha \gamma>0 \\ \Rightarrow & b<0 \end{matrix} \Rightarrow-b>0 $

So, the cubic polynomial $x^{3}+a x^{2}-b x+c$ has all three zeroes which are positive only when all constants $a, b$ and $c$ are negative.

(vii) False, let

$ f(x)=k x^{2}+x+k $

For equal roots. Its discriminant should be zero i.e., $D=b^{2}-4 a c=0$

$ \begin{aligned} \Rightarrow & 1-4 k \cdot k & =0 \\ \Rightarrow & k & = \pm \dfrac{1}{2} \end{aligned} $

So, for two values of $k$, given quadratic polynomial has equal zeroes.

Thinking Process

Firstly, we use the factorisation method i.e., splitting the middle term of quadratic polynomial and find its zeroes. After that use the formula of sum of zeroes and product of zeroes for verification.

Solution

Let

$ \begin{aligned} f(x) & =4 x^{2}-3 x-1 \\ & =4 x^{2}-4 x+x-1 \quad \quad \text{ [by splitting the middle term] } \\ & =4 x(x-1)+1(x-1) \\ & =(x-1)(4 x+1) \end{aligned} $

So, the value of $4 x^{2}-3 x-1$ is zero when $x-1=0$ or $4 x+1=0$ i.e., when $x=1$ or $x=-\dfrac{1}{4}$.

So, the zeroes of $4 x^{2}-3 x-1$ are 1 and $-\dfrac{1}{4}$.

$ \begin{aligned} \therefore \quad \text{ Sum of zeroes } & =1-\dfrac{1}{4}=\dfrac{3}{4}=\dfrac{-(-3)}{4} \\ & =(-1) \dfrac{\text{ Coefficient of } x}{\text{ Coefficient of } x^{2}} \end{aligned} $

and $\quad$ product of zeroes $=(1) \quad-\dfrac{1}{4}=-\dfrac{1}{4}$

$ =(-1)^{2} \cdot \dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{2}} $

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

Solution

Let

$ \begin{aligned} f(x) & =3 x^{2}+4 x-4 \\ & =3 x^{2}+6 x-2 x-4 \quad \quad \text{ [by splitting the middle term] } \\ & =3 x(x+2)-2(x+2) \\ & =(x+2)(3 x-2) \end{aligned} $

So, the value of $3 x^{2}+4 x-4$ is zero when $x+2=0$ or $3 x-2=0$, i.e., when $x=-2$ or $x=\dfrac{2}{3}$. So, the zeroes of $3 x^{2}+4 x-4$ are -2 and $\dfrac{2}{3}$.

$ \begin{aligned} \therefore \quad \text{ Sum of zeroes } & =-2+\dfrac{2}{3}=-\dfrac{4}{3} \\ & =(-1) \cdot \dfrac{\text{ Coefficient of } x}{\text{ Coefficient of } x^{2}} \end{aligned} $

and

$ \begin{aligned} \text{ product of zeroes } & =(-2) \quad \dfrac{2}{3}=\dfrac{-4}{3} \\ & =(-1)^{2} \cdot \dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{2}} \end{aligned} $

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

Solution

Let

$ \begin{aligned} f(t) & =5 t^{2}+12 t+7 \\ & =5 t^{2}+7 t+5 t+7 \\ & =t(5 t+7)+1(5 t+7) \\ & =(5 t+7)(t+1) \end{aligned} $

$ =5 t^{2}+7 t+5 t+7 \quad \text{ [by splitting the middle term] } $

So, the value of $5 t^{2}+12 t+7$ is zero when $5 t+7=0$ or $t+1=0$,

i.e., $\quad$ when $t=\dfrac{-7}{5}$ or $t=-1$.

So, the zeroes of $5 t^{2}+12 t+7$ are $-7 / 5$ and -1 .

$ \begin{aligned} \therefore \quad \text{ Sum of zeroes } & =-\dfrac{7}{5}-1=\dfrac{-12}{5} \\ & =(-1) \cdot \dfrac{\text{ Coefficient of } t}{\text{ Coefficient of } t^{2}} \end{aligned} $

and

$ \begin{aligned} \text{ product of zeroes } & =-\dfrac{7}{5}(-1)=\dfrac{7}{5} \\ & =(-1)^{2} \cdot \dfrac{\text{ Constant term }}{\text{ Coefficient of } t^{2}} \end{aligned} $

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

Solution

Let

$ \begin{aligned} f(t) & =t^{3}-2 t^{2}-15 t \\ & =t(t^{2}-2 t-15) \\ & =t(t^{2}-5 t+3 t-15) \quad \quad \text{ [by splitting the middle term] } \\ & =t[t(t-5)+3(t-5)] \\ & =t(t-5)(t+3) \end{aligned} $

So, the value of $t^{3}-2 t^{2}-15 t$ is zero when $t=0$ or $t-5=0$ or $t+3=0$

i.e., $\quad$ when $t=0$ or $t=5$ or $t=-3$.

So, the zeroes of $t^{3}-2 t^{2}-15 t$ are $-3,0$ and 5 .

$ \begin{aligned} \therefore \quad \text{ Sum of zeroes } & =-3+0+5=2=\dfrac{-(-2)}{1} \\ & =(-1) . \dfrac{\text{ Coefficient of } t^{2}}{\text{ Coefficient of } t^{3}} \end{aligned} $

Sum of product of two zeroes at a time

$ \begin{aligned} & =(-3)(0)+(0)(5)+(5)(-3) \\ & =0+0-15=-15 \\ & =(-1)^{2} \cdot \dfrac{\text{ Coefficient of } t}{\text{ Coefficient of } t^{3}} \end{aligned} $

and product of zeroes $=(-3)(0)(5)=0$

$ =(-1)^{3} \dfrac{\text{ Constant term }}{\text{ Coefficient of } t^{3}} $

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

Solution

Let

$ \begin{aligned} f(x) & =2 x^{2}+\dfrac{7}{2} x+\dfrac{3}{4}=8 x^{2}+14 x+3 \\ & =8 x^{2}+12 x+2 x+3 \\ & =4 x(2 x+3)+1(2 x+3) \\ & =(2 x+3)(4 x+1) \end{aligned} $

So, the value of $8 x^{2}+14 x+3$ is zero when $2 x+3=0$ or $4 x+1=0$,

i.e., $\quad$ when $x=-\dfrac{3}{2}$ or $x=-\dfrac{1}{4}$.

So, the zeroes of $8 x^{2}+14 x+3$ are $-\dfrac{3}{2}$ and $-\dfrac{1}{4}$.

$ \begin{aligned} \therefore \quad \text{ Sum of zeroes } & =-\dfrac{3}{2}-\dfrac{1}{4}=-\dfrac{7}{4}=\dfrac{-7}{2 \times 2} \\ & =-\dfrac{(\text{ Coefficient of } x)}{(\text{ Coefficient of } x^{2})} \end{aligned} $

And

$ \begin{aligned} \text{ roduct of zeroes } & =-\dfrac{3}{2} \quad-\dfrac{1}{4}=\dfrac{3}{8}=\dfrac{3}{2 \times 4} \\ & =\dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{2}} \end{aligned} $

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

Solution

Let

$ \begin{aligned} f(x) & =4 x^{2}+5 \sqrt{2} x-3 \\ & =4 x^{2}+6 \sqrt{2} x-\sqrt{2} x-3 \quad \quad \text{ [by splitting the middle term] } \\ & =2 \sqrt{2} x(\sqrt{2} x+3)-1(\sqrt{2} x+3) \\ & =(\sqrt{2} x+3)(2 \sqrt{2} \cdot x-1) \end{aligned} $

So, the value of $4 x^{2}+5 \sqrt{2} x-3$ is zero when $\sqrt{2} x+3=0$ or $2 \sqrt{2} \cdot x-1=0$,

i.e.,

$ \text{ when } x=-\dfrac{3}{\sqrt{2}} \text{ or } x=\dfrac{1}{2 \sqrt{2}} \text{. } $

So, the zeroes of $4 x^{2}+5 \sqrt{2} x-3$ are $-\dfrac{3}{\sqrt{2}}$ and $\dfrac{1}{2 \sqrt{2}}$.

$ \begin{aligned} \therefore \quad \text{ Sum of zeroes } & =-\dfrac{3}{\sqrt{2}}+\dfrac{1}{2 \sqrt{2}} \\ & =-\dfrac{5}{2 \sqrt{2}}=\dfrac{-5 \sqrt{2}}{4} \\ & =-\dfrac{(\text{ Coefficient of } x)}{(\text{ Coefficient of } x^{2})} \end{aligned} $

and

$ \begin{aligned} \text{ product of zeroes } & =-\dfrac{3}{\sqrt{2}} \cdot \dfrac{1}{2 \sqrt{2}}=-\dfrac{3}{4} \\ & =\dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{2}} \end{aligned} $

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

Solution

Let

$ \begin{aligned} f(s) & =2 s^{2}-(1+2 \sqrt{2}) s+\sqrt{2} \\ & =2 s^{2}-s-2 \sqrt{2} s+\sqrt{2} \\ & =s(2 s-1)-\sqrt{2}(2 s-1) \\ & =(2 s-1)(s-\sqrt{2}) \end{aligned} $

So, the value of $2 s^{2}-(1+2 \sqrt{2}) s+\sqrt{2}$ is zero when $2 s-1=0$ or $s-\sqrt{2}=0$,

i.e., $\quad$ when $s=\dfrac{1}{2}$ or $s=\sqrt{2}$.

So, the zeroes of $2 s^{2}-(1+2 \sqrt{2}) s+\sqrt{2}$ are $\dfrac{1}{2}$ and $\sqrt{2}$.

$\therefore \quad$ Sum of zeroes $=\dfrac{1}{2}+\sqrt{2}=\dfrac{1+2 \sqrt{2}}{2}=\dfrac{-[-(1+2 \sqrt{2})]}{2}=\dfrac{(\text{ Coefficient of } s)}{(\text{ Coefficient of } s^{2})}$

and product of zeroes $=\dfrac{1}{2} \cdot \sqrt{2}=\dfrac{1}{\sqrt{2}}=\dfrac{\text{ Constant term }}{\text{ Coefficient of } s^{2}}$

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

Solution

Let

$ \begin{aligned} f(v) & =v^{2}+4 \sqrt{3} v-15 \\ & =v^{2}+(5 \sqrt{3}-\sqrt{3}) v-15 \quad \text{ [by splitting the middle term] } \\ & =v^{2}+5 \sqrt{3} v-\sqrt{3} v-15 \\ & =v(v+5 \sqrt{3})-\sqrt{3}(v+5 \sqrt{3}) \\ & =(v+5 \sqrt{3})(v-\sqrt{3}) \end{aligned} $

So, the value of $v^{2}+4 \sqrt{3} v-15$ is zero when $v+5 \sqrt{3}=0$ or $v-\sqrt{3}=0$,

i.e., $\quad$ when $v=-5 \sqrt{3}$ or $v=\sqrt{3}$.

So, the zeroes of $v^{2}+4 \sqrt{3} v-15$ are $-5 \sqrt{3}$ and $\sqrt{3}$.

$\therefore \quad$ Sum of zeroes $=-5 \sqrt{3}+\sqrt{3}=-4 \sqrt{3}$

$ =(-1) \cdot \dfrac{\text{ Coefficient of } v}{\text{ Coefficient of } v^{2}} $

and $\quad$ product of zeroes $=(-5 \sqrt{3})(\sqrt{3})$

$ \begin{aligned} & =-5 \times 3=-15 \\ & =(-1)^{2} \cdot \dfrac{\text{ Constant term }}{\text{ Coefficient of } v^{2}} \end{aligned} $

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

Solution

Let

$ \begin{aligned} f(y)=y^{2}+\dfrac{3}{2} \sqrt{5} y-5 & =2 y^{2}+3 \sqrt{5} y-10 \\ & =2 y^{2}+4 \sqrt{5} y-\sqrt{5} y-10 \quad \text{ [by splitting the middle term] } \\ & =2 y(y+2 \sqrt{5})-\sqrt{5}(y+2 \sqrt{5}) \\ & =(y+2 \sqrt{5})(2 y-\sqrt{5}) \end{aligned} $

So, the value of $y^{2}+\dfrac{3}{2} \sqrt{5} y-5$ is zero when $(y+2 \sqrt{5})=0$ or $(2 y-\sqrt{5})=0$, i.e.,

$ \text{ when } y=-2 \sqrt{5} \text{ or } y=\dfrac{\sqrt{5}}{2} \text{. } $

So, the zeroes of $2 y^{2}+3 \sqrt{5} y-10$ are $-2 \sqrt{5}$ and $\dfrac{\sqrt{5}}{2}$.

$\therefore \quad$ Sum of zeroes $=-2 \sqrt{5}+\dfrac{\sqrt{5}}{2}=\dfrac{-3 \sqrt{5}}{2}=-\dfrac{(\text{ Coefficient of } y)}{(\text{ Coefficient of } y^{2})}$

And product of zeroes $=-2 \sqrt{5} \times \dfrac{\sqrt{5}}{2}=-5=\dfrac{\text{ Constant term }}{\text{ Coefficient of } y^{2}}$

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

Solution

Let

$ \begin{aligned} f(y) & =7 y^{2}-\dfrac{11}{3} y-\dfrac{2}{3} \\ & =21 y^{2}-11 y-2 \\ & =21 y^{2}-14 y+3 y-2 \quad \quad \text{ [by splitting the middle term] } \\ & =7 y(3 y-2)+1(3 y-2) \\ & =(3 y-2)(7 y+1) \end{aligned} $

So, the value of $7 y^{2}-\dfrac{11}{3} y-\dfrac{2}{3}$ is zero when $3 y-2=0$ or $7 y+1=0$,

i.e., $\quad$ when $y=\dfrac{2}{3}$ or $y=-\dfrac{1}{7}$.

So, the zeroes of $7 y^{2}-\dfrac{11}{3} y-\dfrac{2}{3}$ are $\dfrac{2}{3}$ and $-\dfrac{1}{7}$.

$\therefore \quad$ Sum of zeroes $=\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{14-3}{21}=\dfrac{11}{21}=-\dfrac{-11}{3 \times 7}$

$ =(-1) \cdot \dfrac{\text{ Coefficient of } y}{\text{ Coefficient of } y^{2}} $

and product of zeroes $=\dfrac{2}{3} \quad-\dfrac{1}{7}=\dfrac{-2}{21}=\dfrac{-2}{3 \times 7}$

$ =(-1)^{2} \cdot \dfrac{\text{ Constant term }}{\text{ Coefficient of } y^{2}} $

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

Thinking Process

(i) Firstly we use the concept or method of formation a quadratic equation, i.e., $f(x)=x^{2}$-(sum of the zeroes) $x+$ (product of zeroes).

(ii) After making a quadratic polynomial we factorise it by splitting the middle term and get the required zeroes.

Solution

(i) Given that, sum of zeroes $(S)=-\dfrac{8}{3}$

and product of zeroes $(P)=\dfrac{4}{3}$

$\therefore$ Required quadratic expression, $f(x)=x^{2}-S x+P$

$ =x^{2}+\dfrac{8}{3} x+\dfrac{4}{3}=3 x^{2}+8 x+4 $

Using factorisation method, $=3 x^{2}+6 x+2 x+4$

$ =3 x(x+2)+2(x+2)=(x+2)(3 x+2) $

Hence, the zeroes of $f(x)$ are -2 and $-\dfrac{2}{3}$.

(ii) Given that, $S=\dfrac{21}{8}$ and $P=\dfrac{5}{16}$

$\therefore$ Required quadratic expression, $f(x)=x^{2}-S x+P$

$ =x^{2}-\dfrac{21}{8} x+\dfrac{5}{16}=16 x^{2}-42 x+5 $

Using factorisation method $=16 x^{2}-40 x-2 x+5$

$ =8 x(2 x-5)-1(2 x-5)=(2 x-5)(8 x-1) $

Hence, the zeroes of $f(x)$ are $\dfrac{5}{2}$ and $\dfrac{1}{8}$

(iii) Given that, $S=-2 \sqrt{3}$ and $P=-9$

$\therefore$ Required quadratic expression,

$ \begin{aligned} f(x) & =x^{2}-S x+P=x^{2}+2 \sqrt{3} x-9 \\ & =x^{2}+3 \sqrt{3} x-\sqrt{3} x-9 \quad \text{ [Using factorisation method] } \\ & =x(x+3 \sqrt{3})-\sqrt{3}(x+3 \sqrt{3}) \\ & =(x+3 \sqrt{3})(x-\sqrt{3}) \end{aligned} $

Hence, the zeroes of $f(x)$ are $-3 \sqrt{3}$ and $\sqrt{3}$.

(iv) Given that, $S=-\dfrac{3}{2 \sqrt{5}}$ and $P=-\dfrac{1}{2}$

$\therefore$ Required quadratic expression,

$ \begin{aligned} f(x) & =x^{2}-S x+P=x^{2}+\dfrac{3}{2 \sqrt{5}} x-\dfrac{1}{2} \\ & =2 \sqrt{5} x^{2}+3 x-\sqrt{5} \end{aligned} $

Using factorisation method, $=2 \sqrt{5} x^{2}+5 x-2 x-\sqrt{5}$

$ =\sqrt{5} x(2 x+\sqrt{5})-1(2 x+\sqrt{5}) $

$ =(2 x+\sqrt{5})(\sqrt{5} x-1) $

Hence, the zeroes of $f(x)$ are $-\dfrac{\sqrt{5}}{2}$ and $\dfrac{1}{\sqrt{5}}$.

Thinking Process

Using the following relations related to a cubic polynomial.

(i) Sum of the zeroes $=(-1) \cdot \dfrac{\text{ Coefficient of } x^{2}}{\text{ Coefficient of } x^{3}}$

(ii) Sum of product of two zeroes at a time $=(-1)^{2} \cdot \dfrac{\text{ Coefficient of } x}{\text{ Coefficient of } x^{3}}$

Solution

Let

$ f(x)=x^{3}-6 x^{2}+3 x+10 $

Given that, $a,(a+b)$ and $(a+2 b)$ are the zeroes of $f(x)$. Then,

$ \begin{aligned} & \text{ Sum of the zeroes }=-\dfrac{(\text{ Coefficient of } x^{2})}{(\text{ Coefficient of } x^{3})} \\ & \Rightarrow \quad a+(a+b)+(a+2 b)=-\dfrac{(-6)}{1} \\ & \Rightarrow \quad 3 a+3 b=6 \\ & \Rightarrow \quad a+b=2 \\ & \Rightarrow \quad a(a+b)+(a+b)(a+2 b)+a(a+2 b)=\dfrac{3}{1} \\ & \Rightarrow a(a+b)+(a+b){(a+b)+b}+a{(a+b)+b}=3 \\ & \Rightarrow \quad 2 a+2(2+b)+a(2+b)=3 \\ & \Rightarrow \quad 2 a+2(2+2-a)+a(2+2-a)=3 \\ & \Rightarrow \quad 2 a+8-2 a+4 a-a^{2}=3 \\ & \Rightarrow \quad-a^{2}+8=3-4 a \\ & \Rightarrow \quad a^{2}-4 a-5=0 \end{aligned} $

Using factorisation method,

$ \begin{aligned} & \Rightarrow \quad a(a-5)+1(a-5)=0 \\ & \Rightarrow \quad(a-5)(a+1)=0 \\ & \Rightarrow \quad a=-1,5 \\ & a=-1 \text{ and } b=3 \\ & a,(a+b),(a+2)=-1,(-1+3),(-1+6) \text{ or }-1,2,5 \\ & a=5 \text{ and } b=-3, \text{ then } \end{aligned} $

[using Eq. (i)]

$ a,(a+b),(a+2 b)=5,(5-3),(5-6) \text{ or } 5,2,-1 \text{. } $

Hence, the required values of $a$ and $b$ are $a=-1$ and $b=3$ or $a=5, b=-3$ and the zeroes are $-1,2$ and 5 .

Thinking Process

Use division algorithm and get a quadratic polynomial. Further factorize the quadratic polynomial by factorisation method and get the other two required roots.

Solution

Let $f(x)=6 x^{3}+\sqrt{2} x^{2}-10 x-4 \sqrt{2}$ and given that, $\sqrt{2}$ is one of the zeroes of $f(x)$ i.e., $(x-\sqrt{2})$ is one of the factor of given cubic polynomial.

Now, using division algorithm,

$\therefore \quad 6 x^{3}+\sqrt{2} x^{2}-10 x-4 \sqrt{2}=(6 x^{2}+7 \sqrt{2} x+4) \times(x-\sqrt{2})+0$

$[\because$ dividend $=$ divisor $\times$ quotient + remainder $]$

$ \begin{aligned} & =(x-\sqrt{2})(6 x^{2}+4 \sqrt{2} x+3 \sqrt{2} x+4) \\ & =(x-\sqrt{2}){\sqrt{2} x(3 \sqrt{2} x+4)+1(3 \sqrt{2} x+4)} \\ & =(x-\sqrt{2}){(3 \sqrt{2} x+4)(\sqrt{2} x+1)} \\ & =(x-\sqrt{2})(\sqrt{2} x+1)(3 \sqrt{2} x+4) \end{aligned} $

So, its other zeroes are $-\dfrac{1}{\sqrt{2}}$ and $-\dfrac{4}{3 \sqrt{2}}$.

Solution

Given that, $x^{2}+2 x+k$ is a factor of $2 x^{4}+x^{3}-14 x^{2}+5 x+6$, then we apply division algorithm,

Since, $(x^{2}+2 x+k)$ is a factor of $2 x^{4}+x^{3}-14 x^{2}+5 x+6$.

So, when we apply division algorithm remainder should be zero.

$ \begin{aligned} & \therefore \quad(7 k+21) x+(2 k^{2}+8 k+6)=0 \cdot x+0 \\ & \Rightarrow \quad 7 k+21=0 \text{ and } 2 k^{2}+8 k+6=0 \\ & \Rightarrow \quad k=-3 \text{ or } k^{2}+4 k+3=0 \\ & \Rightarrow \quad k(k+3)+1(k+3)=0 \\ & \Rightarrow \quad(k+1)(k+3)=0 \\ & \Rightarrow \quad k=-1 \text{ or }-3 \end{aligned} $

$ \Rightarrow \quad k^{2}+3 k+k+3=0 \quad \text{ [by splitting middle term] } $

Here, if we take $k=-3$, then remainder will be zero.

Thus, the required value of $k$ is -3 .

$ \begin{matrix} \text{ Now, } & \text{ Dividend }=\text{ Divisor } \times \text{ Quotient }+ \text{ Remainder } \\ \Rightarrow \quad & 2 x^{4}+x^{3}-14 x^{2}+5 x+16=(x^{2}+2 x-3)(2 x^{2}-3 x-2) \end{matrix} $

Using factorisation method,

$ \begin{aligned} & =(x^{2}+3 x-x-3)(2 x^{2}-4 x+x-2)[\text{ by splitting middle term }] \\ & ={x(x+3)-1(x+3)}{2 x(x-2)+1(x-2)} \\ & =(x-1)(x+3)(x-2)(2 x+1) \end{aligned} $

Hence, the zeroes of $x^{2}+2 x-3$ are $1,-3$ and the zeroes of $2 x^{4}+x^{3}-14 x^{2}+5 x+6$ are $1,-3,2, \dfrac{-1}{2}$.

Solution

Let $f(x)=x^{3}-3 \sqrt{5} x^{2}+13 x-3 \sqrt{5}$ and given that, $(x-\sqrt{5})$ is a one of the factor of $f(x)$. Now, using division algorithm,

$$ \begin{aligned} & \therefore \quad x^{3}-3 \sqrt{5} x^{2}+13 x-3 \sqrt{5}=(x^{2}-2 \sqrt{5} x+3) \times(x-\sqrt{5}) \\ & {[\because \text{ dividend }=\text{ divisor } \times \text{ quotient }+ \text{ remainder }]} \\ & =(x-\sqrt{5})[x^{2}-{(\sqrt{5}+\sqrt{2})+(\sqrt{5}-\sqrt{2})} x+3] \quad[\text{ by splitting the middle term }] \\ & =(x-\sqrt{5})[x^{2}-(\sqrt{5}+\sqrt{2}) x-(\sqrt{5}-\sqrt{2}) x+(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})] \\ & {[\because 3=(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})]} \\ & =(x-\sqrt{5})[x{x-(\sqrt{5}+\sqrt{2})}-(\sqrt{5}-\sqrt{2}){x-(\sqrt{5}+\sqrt{2})}] \\ & =(x-\sqrt{5}){x-(\sqrt{5}+\sqrt{2})}{x-(\sqrt{5}-\sqrt{2})} \end{aligned} $$

Hence, all the zeroes of polynomial are $\sqrt{5},(\sqrt{5}+\sqrt{2})$ and $(\sqrt{5}-\sqrt{2})$.

Thinking Process

(i) Firstly, we use the division algorithm to get the remainder. Since, $q(x)$ is a factor of $p(x)$. So, remainder should be zero.

(ii) Now, equating the like terms of $x$ and get the values of ’ $a$ ’ and ’ $b$ ‘. After that factorise the quotient by splitting the middle term method.

(iii) Finally, we get two more zeroes, which are not the zeroes of $p(x)$.

Solution

Given that the zeroes of $q(x)=x^{3}+2 x^{2}+a$ are also the zeroes of the polynomial $p(x)=x^{5}-x^{4}-4 x^{3}+3 x^{2}+3 x+b$ i.e., $q(x)$ is a factor of $p(x)$. Then, we use a division algorithm.

$ \begin{aligned} & x^{2}-3 x+2 \\ & \dfrac{x^{5}+2 x^{4}+a x^{2}}{-3 x^{4}-4 x^{3}+(3-a) x^{2}+3 x+b} \\ & \quad \dfrac{-3 x^{4}-6 x^{3}-3 a x}{x^{5}-x^{4}-4 x^{3}+3 x^{2}+3 x+b} \\ & \dfrac{2 x^{3}+(3-a) x^{2}+(3+3 a) x+b}{-(1+a) x^{2}+(3+3 a) x+(b-2 a)} \end{aligned} $

If $(x^{3}+2 x^{2}+a)$ is a factor of $(x^{5}-x^{4}-4 x^{3}+3 x^{2}+3 x+b)$, then remainder should be zero.

$ \text{ i.e., } \quad \begin{matrix} -(1+a) x^{2}+(3+3 a) x+(b-2 a)=0 \\ =0 \cdot x^{2}+0 \cdot x+0 \end{matrix} $

On comparing the coefficient of $x$, we get

$ \begin{aligned} & a+1=0 \\ & \Rightarrow \quad a=-1 \\ & \text{ and } \quad b-2 a=0 \\ & \Rightarrow \quad b=2 a \\ & \therefore \quad q(x)=x^{3}+2 x^{2}-1 \\ & p(x)=(x^{3}+2 x^{2}-1)(x^{2}-3 x+2)+0 \\ & =(x^{3}+2 x^{2}-1){x^{2}-2 x-x+2} \\ & =(x^{3}+2 x^{2}-1)(x-2)(x-1) \end{aligned} $

$ \Rightarrow \quad b=2(-1)=-2 \quad[\because a=-1] $

For $a=-1$ and $b=-2$, the zeroes of $q(x)$ are also the zeroes of the polynomial $p(x)$.

$ p(x)=x^{5}-x^{4}-4 x^{3}+3 x^{2}+3 x-2 $

Hence, the zeroes of $p(x)$ are 1 and 2 which are not the zeroes of $q(x)$.