POLYNOMIALS

POLYNOMIALS :

An algebraic expression $f(x)$ of the form $(fx)=a_0+a_{1}x+a_2{x}^2+\dots \dots .+a_n{x}^n$, where $a_0,a_1,a_2\dots \dots ..a_n$ are real numbers and all the index of $\mathbf{x}$ are non-negative integers is called polynomials in $\mathbf{x}$ and the highest Index $\mathbf{n}$ in called the degree of the polynomial, if $a_n\ne 0$.

POLYNOMIALS :

4.1 (a) Zero Degree Polynomial :

Any non-zero number is regarded as a polynomial of degree zero or zero degree polynomial. For example, $\mathbf{f}(\mathbf{x})=\mathbf{a}$, where $a\ne 0$ is a zero degree polynomial, since we can write $\mathbf{f}(\mathbf{x})=\mathbf{a}$ as $\mathbf{f}(\mathbf{x})=\mathbf{a}{\mathbf{x}}^0$.

POLYNOMIALS :

4.1 (b) Constant Polynomial :

A polynomial of degree zero is called a constant polynomial. For example, $f(x)=7$.

POLYNOMIALS

4.1 (c) Linear Polynomial :

A polynomial of degree 1 is called a linear polynomial.

For example: $p(x)=4x-3$ and $f(t)=\sqrt{3}t+5$ are linear polynomials.

POLYNOMIALS :

4.1 (d) Quadratic Polynomial :

A polynomial of degree 2 is called quadratic polynomial.

For example: $f(x)=2x^2+5x-\frac{3}{5}$ and $g(y)=3y^2-5$ are quadratic polynomials with real coefficients.

POLYNOMIALS

IMPORTANT FORMULAE :

$(x+a{)}^2=x^2+2ax+a^2$

$(x-a{)}^2=x^2-2ax+a^2$

$x^2-a^2=(x+a)(x-a)$

$x^3+a^3=(x+a)(x^2-ax+a^2)=(x+a{)}^3-3xa(x+a)$

$x^3-a^3=(x-a)(x^2+ax+a^2)=(x-a{)}^3+3xa(x-a)$

$(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$

$(a+b{)}^3=a^3+b^3+3ab(a+b)$

$(a-b{)}^3=a^3-b^3-3ab(a-b)$

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Special Case: If $a+b+c=0$ then $a^3+b^3+c^3=3abc$.

POLYNOMIALS

4.2 GRAPH OF POLYNOMIALS :

In algebraic or in set theoretic language the graph of a polynomial $f(x)$ is the collection (or set) of all points $(\mathbf{x},\mathbf{y})$, where $\mathbf{y}=\mathbf{f}(\mathbf{x})$. In geometrical or in graphical language the graph of a polynomial $f(x)$ is a smooth free hand curve passing through points $.{\mathbf{x}}_1,{\mathbf{y}}_1),({\mathbf{x}}_2,{\mathbf{y}}_2),({\mathbf{x}}_3,{\mathbf{y}}_3),\dots .$. etc. where ${\mathbf{y}}_1,{\mathbf{y}}_2,{\mathbf{y}}_3,\dots$ are the values of the polynomial $\mathbf{f}(\mathbf{x})$ at ${\mathbf{x}}_1,{\mathbf{x}}_2,{\mathbf{x}}_3,\dots$ respectively.

In order to draw the graph of a polynomial $\mathbf{f}(\mathbf{x})$, follow the following algorithm.

POLYNOMIALS

ALGORITHM :

Step (i) Find the values ${\mathbf{y}}_1,{\mathbf{y}}_2,\dots ..{\mathbf{y}}_{\mathbf{n}}$ of polynomial $f(x)$ on different points ${\mathbf{x}}_1,{\mathbf{x}}_2,\dots \dots .{\mathbf{x}}_n\dots \dots \dots \dots$ and prepare a table that gives values of $\mathbf{y}$ or $\mathbf{f}(\mathbf{x})$ for various values of $\mathbf{x}$.

Step (ii) Plot that points $(x_1,y_1),(x_2,y_2),(x_3,y_3),\dots ..(x_n,y_n)\dots$. on rectangular co-ordinate system. In plotting these points use different scales on the $X$ and $Y$ axes.

Step (iii) Draw a free hand smooth curve passing through points plotted in step 2 to get the graph of the polynomial $\mathbf{f}(\mathbf{x})$.

POLYNOMIALS

4.2 (a) Graph of a Linear Polynomial :

Consider a linear polynomial $f(x)=ax+b,a\ne 0$ Graph of $\mathbf{y}=\mathbf{a}\mathbf{x}+\mathbf{b}$ is a straight line. That in why $f(x)=$ $ax+b$ is called a linear polynomial. Since two points determine a straight line, so only two points need to plotted to draw the line $\mathbf{y}=\mathbf{a}\mathbf{x}+\mathbf{b}$. The line represented by $\mathbf{y}=\mathbf{a}\mathbf{x}+\mathbf{b}$ crosses the $X$-axis at exactly one point, namely $(-\frac{b}{a},0)$.

POLYNOMIALS

Ex. 1 Draw the graph of the polynomial $f(x)=2x-5$. Also, find the coordinates of the point where it crosses $X$ axis.

POLYNOMIALS

Sol. Let $y=2x-5$.

The following table list the values of $\mathbf{y}$ corresponding to different values of $\mathbf{x}$.

The points A $(1,-3)$ and $B(4,3)$ are plotted on the graph paper on a suitable scale. A line is drawn passing through these points to obtain the graphs of the given polynomial.

POLYNOMIALS

4.2 (b) Graph of a Quadratic Polynomial :

Let $a,b,c$ be real numbers and $a\ne 0$. Then $f(x)=a{x}^2+bx+c$ is known as a quadratic polynomial in $x$. Graph of the quadratic polynomial i.e. he curve whose equation is $y=a{x}^2+bx+c,a\ne 0$ Graph of $a$ quadratic polynomial is always a parabola.

Let $y=a{x}^2+bx+c$, where $a\ne 0$

$\Rightarrow 4ay=4a^2{x}^2+4abx+4ac$

$\Rightarrow 4ay=4a^2{x}^2+4abx+b^2-b^2+4ac$

$\Rightarrow 4ay=(2ax+b{)}^2-(b^2-4ac)$

$\Rightarrow 4ay+(b^2-4ac)=(2ax+b{)}^2\Rightarrow 4ay+(b^2-4ac)=4a^2(x+b/2a{)}^2$

$\Rightarrow 4ay+\frac{b^2-4ac}{4a}=4a^2(x+\frac{b}{2a}{)}^2$

$\Rightarrow (y+\frac{D}{4a})=a(a+\frac{b}{2a}{)}^2$

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where $\mathbf{D}={\mathbf{b}}^2$ - $\mathbf{4}\mathbf{a}\mathbf{c}$ is the discriminate of the quadratic equation.

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REMARKS :

Shifting the origin at $(-\frac{b}{2a},-\frac{D}{4a})$, we have $X=x-(-\frac{b}{2a})$ and $Y=y-\frac{(-D)}{4a}$

Substituting these values in (i), we obtain

$Y=a{X}^2$ ….(ii)

which is the standard equation of parabola

Clearly, this is the equation of a parabola having its vertex at $(-\frac{b}{2a},\frac{D}{4a})$.

The parabola opens upwards or downwards according as a $>0$ or $a<0$.

POLYNOMIALS

4.3 SIGN OF QUADRTIV EXPRESSIONS :

Let $\alpha$ be a real root of $\mathbf{a}{\mathbf{x}}^2+\mathbf{b}\mathbf{x}+\mathbf{c}=\mathbf{0}$. Then $a{\alpha }^2+b\alpha +c=0$ Point $(\alpha ,0)$ lies on $y=ax{x}^2+bx+c$. Thus, every real root of $a{x}^2+bx+c=0$ represents a point of intersecting of the parabola with the $X$-axis.

Conversely, if the parabola $y=a{x}^2+bx+c$ intersects the $X$-axis at a point $(\alpha ,0)$ then $(\alpha ,0)$ satisfies the equation $y=a{x}^2+bx+c$

$\Rightarrow a{\alpha }^2+b\alpha +c=0[\alpha .$ is a real root of $a{x}^2+bx+c=0$ ]

Thus, the intersection of the parabola $y=a{x}^2+bx+c$ with $X$-axis gives all the real roots of $a{x}^2+bx+c=$ 0 . Following conclusions may be drawn :-

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(i) If $D>0$, the parabola will intersect the $x$-axis in two distinct points and vice-versa.

The parabola meets $x$-axis at $\alpha =\frac{-b-\sqrt{D}}{2a}$ and $\beta =\frac{-b+\sqrt{D}}{2a}$

Roots are real & distinct

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(ii) If $D=0$, the parabola will just touch the $x$-axis at one point and vice-versa.

Roots are equal

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(iii) If $D<0$, the parabola will not intersect $x$-axis at all and vice-versa.

Roots are imaginary

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REMARKS

$\star \mathrm{\forall }x\in R,y>0$ only if $a>0$ & $D\equiv b^2-4ac<0$

$\star \mathrm{\forall }x\in R,y<0$ only if $a<0$ & $D\equiv b^2-4ac<0$

POLYNOMIALS

Ex. 2 Draw the graph of the polynomial $f(x)=x^2-2x-8$

POLYNOMIALS

Sol. Let $y=x^2-2x-8$.

The following table gives the values of $y$ or $f(x)$ for various values of $x$.

POLYNOMIALS

Let us plot the points $(-4,16),(-3,7),(-2,0),(-1,-5),(0,-8),(1,-9),(2,-8),(3,-5),(4,0),(5,7)$ and $(6,16)$ on a graphs paper and draw a smooth free hand curve passing through these points. The curve thus obtained represents the graphs of the polynomial $f(x)=x^2-2x-8$. This is called a parabola. The lowest point $P$, called a minimum points, is the vertex of the parabola. Vertical line passing through $P$ is called the axis of the parabola. Parabola is symmetric about the axis. So, it is also called the line of symmetry.

POLYNOMIALS

Observations :

From the graphs of the polynomial $f(x)=x^2-2x-8$, following observations can be drawn :

(i) The coefficient of $x^2$ in $f(x)=x^2-2x-8$ is 1 (a positive real number) and so the parabola opens upwards.

(ii) $D=b^2-4ac=4+32=36>0$. So, the parabola cuts $X$-axis at two distinct points.

(iii) On comparing the polynomial $x^2-2x-8$ with $a{x}^2+bx+c$, we get $a=1,b=-2$ and $c=-8$.

The vertex of the parabola has coordinates $(1,-9)$ i.e. $(\frac{-b}{2a},\frac{-D}{4a})$, where $D\equiv b^2-4ac$.

(iv) The polynomial $f(x)=x^2-2x-8=(x-4)(x+2)$ is factorizable into two distinct linear factors $(x-4)$ and $(x+2)$. So, the parabola cuts $X$-axis at two distinct points $(4,0)$ and $(-2,0)$. the $x$-coordinates of these points are zeros of $f(x)$.

POLYNOMIALS

Ex. 3 Draw the graphs of the quadratic polynomial $f(x)=3-2x-x^2$.

POLYNOMIALS

Sol. Let $y=f(x)$ or, $y=3-2x-x^2$.

Let us list a few values of $y=3-2x-x^2$ corresponding to a few values of $x$ as follows :

POLYNOMIALS

Thus, the following points lie on the graph of the polynomial $y=2-2x-x^2$ :

$(-5,-12),(-4,-5),(-3,0),(-2,4),(-1,4),(0,3),(1$, $0),(2,-5),(3,-12)$ and $(4,-21)$.

Let plot these points on a graph paper and draw a smooth free hand curve passing through these points to obtain the graphs of $y$ $=3-2x-x^2$. The curve thus obtained represents a parabola, as shown in figure. The highest point $P(-1,4)$, is called a maximum points, is the vertex of the parabola. Vertical line through $P$ is the axis of the parabola. Clearly, parabola is symmetric about the axis.

POLYNOMIALS

Observations:Following observations from the graph of the polynomial $f(x)=3-2x-x^2$ is as follows :

(i) The coefficient of $x^2$ in $f(x)=3-2x-x^2$ is -1 i.e. a negative real number and so the parabola opens downwards.

(ii) $D\equiv b^2-4ax=4+12=16>0$. So, the parabola cuts $x$-axis two distinct points.

(iii) On comparing the polynomial $3-2x-x^2$ with $a{x}^2+bc+c$, we have $a=-1,b=-2$ and $c=3$. The vertex of the parabola is at the point $(-1,4)$ i.e. at $(\frac{-b}{2a},\frac{-D}{4a})$, where $D=b^2-4ac$.

(iv) The polynomial $f(x)=3-2x-x^2=(1-x)(x+3)$ is factorizable into two distinct linear factors $(1-x)$ and $(x+3)$. So, the parabola cuts $X$-axis at two distinct points $(1,0)$ and $(-3,0)$. The co-ordinates of these points are zeros of $f(x)$.

POLYNOMIALS

4.4 GRAPH OF A CUBIC POLYNOMIAL :

Graphs of a cubic polynomial does not have a fixed standard shape. Cubic polynomial graphs will always cross $X$-axis at least once and at most thrice.

POLYNOMIALS

Ex. 4 Draw the graphs of the polynomial $f(x)=x^3-4x$.

POLYNOMIALS

Sol. Let $y=f(x)$ or, $y=x^2-4x$.

The values of $y$ for variable value of $x$ are listed in the following table :

POLYNOMIALS

Thus, the curve $y=x^3-4x$ passes through the points $(-3,-15),(-2,0),(-1,3),(0,0),(1,-3),(2,0),(3,15),(4,48)$ etc.Plotting these points on a graph paper and drawing a free hand smooth curve through these points, we obtain the graph of the given polynomial as shown figure.

Observations :

For the graphs of the polynomial $f(x)=x^3-4x$, following observations are as follows :-

(i) The polynomial $f(x)=x^3-4x=x(x^2-4)=x(x-2)(x+2)$ is factorizable into three distinct linear factors. The curve $y=f(x)$ also cuts $X$-axis at three distinct points.

(ii) We have, $f(x)=x(x-2)(x+2)$ Therefore 0,2 and -2 are three zeros of $f(x)$. The curve $y$ $=f(x)$ cuts $X$-axis at three points $O(0,0),P(2,0)$ and $Q(-2,0)$.

POLYNOMIALS

4.5 RELATIONSHIP BETWEEN ZEROS AND COEFFICIENTS OF A QUADRATIC POLYNOMIAL :

Let $\alpha$ and $\beta$ be the zeros of a quadratic polynomial $f(x)=a{x}^2+bx+c$. By facto $r$ theorem $(x-\alpha )$ and $(x-\beta )$ are the factors of $f(x)$.

$\therefore f(x)=k(x-\alpha )(x-\beta )$ are the factors of $f(x)$

$\Rightarrow a{x}^2+bx+c=k{x}^2-(\alpha +\beta )x+\alpha \beta$

$\Rightarrow a{x}^2+bx+c=k{x}^2-k(\alpha +\beta )x+k\alpha \beta$

Comparing the coefficients of $x^2,x$ and constant terms on both sides, we get $a=k,b=-k(\alpha +\beta )$ and $k\alpha \beta$

$\Rightarrow \alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}\Rightarrow \alpha +\beta =-\frac{\text{ Coefficient of }x}{\text{ Coefficient of }{x}^2}$ and $\alpha \beta =\frac{\text{ Constant term }}{\text{ Coefficient of }{x}^2}$

Hence,

Sum of the zeros $=-\frac{b}{a}=-\frac{\text{ Coefficient of }x}{\text{ Coefficient of }{x}^2}$

Product of the zeros $=\frac{c}{a}=\frac{\text{ Constant term }}{\text{ Coefficient of }{x}^2}$

POLYNOMIALS

REMAKRS :

If $\alpha$ and $\beta$ are the zeros of a quadratic polynomial $f(x)$. The , the polynomial $f(x)$ is given by

$f(x)=k{x}^2-(\alpha +\beta )x+\alpha \beta$

or $f(x)=k${$x^2-$( Sum of the zeros) $x+$ Product of the zeros }

POLYNOMIALS

Ex. 5 Find the zeros of the quadratic polynomial $f(x)=x^2-2x-8$ and verify and the relationship between the zeros and their coefficients.

POLYNOMIALS

Sol. $f(x)=x^2-2x-8$

$.\Rightarrow f(x)=x^2-4x+2x-8\Rightarrow f(x)=x(x-4)+2(x-4)]$

$\Rightarrow f(x)=(x-4)(x+2)$

Zeros of $f(x)$ are given by $f(x)=0$

$\Rightarrow x^2-2x-8=0\Rightarrow (x-4)(x+2)=0$

$\Rightarrow x=4$ or $x=-2$

So, $\alpha =4$ and $\beta =-2$

$\therefore$ sum of zeros $\alpha +\beta$

$=4-2=2$

POLYNOMIALS

Also, sum of zeros $=-\frac{\text{ Coefficient of }x}{\text{ Coefficient of }{x}^2}=\frac{-(-2)}{1}=2$

So, sum of zeros $=\alpha +\beta =-\frac{\text{ Coefficient of }x}{\text{ Coefficient of }{x}^2}$

Now, product of zeros $=\alpha \beta =(4)(-2)=-8$

Also, product of zeros $=\frac{\text{ Constan }t\text{ term }}{\text{ Coefficient of }{x}^2}=\frac{-8}{1}=-8$

$\therefore$ Product of zeros $=\frac{\text{ Constant term }}{\text{ Coefficient of }{x}^2}=\alpha \beta$.

POLYNOMIALS

Ex. 6 Find a quadratic polynomial whose zeros are $5+\sqrt{2}$ and $5-\sqrt{2}$

POLYNOMIALS

Sol. Given $\alpha =5+\sqrt{2},\beta =5-\sqrt{2}$

$\therefore f(x)=k{x}^2-x(\alpha +\beta )+\alpha \beta$ Here, $\alpha +\beta =5+\sqrt{2}+5-\sqrt{2}=10$

and $\alpha \beta =(5+\sqrt{2})(5-\sqrt{2})=25-2=23$

$\therefore f(x)=k{x}^2-10x+23$, where, $k$ is any non-zero real number.

POLYNOMIALS

Ex. 7 Sum of product of zeros of quadratic polynomial are 5 and 17 respectively. Find the polynomial.

POLYNOMIALS

Sol. Given : Sum of zeros $=5$ and product of zeros $=17$

So, quadratic polynomial is given by

$\Rightarrow f(x)=k${$x^2-x$ (sum of zeros) + product of zeros }

$\Rightarrow f(x)=k{x}^2-5x+17$, where, $k$ is any non-zero real number,

POLYNOMIALS

4.6 RELATIONSHIP BETWEEN ZEROS AND COEFFICIENTS OF A CUBIC POLYNOMIAL :

Let $\alpha ,\beta ,\gamma$ be the zeros of a cubic polynomial $f(x)=a{x}^3+b{x}^2+cx+d,a\ne 0$ Then, by factor theorem, $a-\alpha ,x-\beta$ and $x-\gamma$ are factors of $f(x)$. Also, $f(x)$ being a cubic polynomial cannot have more than three linear factors.

$\begin{array}{rl}& \therefore f(x)=k(x-\alpha )(x-\beta )(x-\gamma )\Rightarrow a{x}^3+b{x}^2+cx+d=k(x-\alpha )(x-\beta )(x-\gamma )\\ & \Rightarrow a{x}^3+b{x}^2+cx+d=k{x}^3-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +\gamma \alpha )x-\alpha \beta \gamma \end{array}$

$\Rightarrow a{x}^3+b{x}^2+cx+d=kx{x}^3-k(\alpha +\beta +\gamma )x^2+k(\alpha \beta +\beta \gamma +\gamma \alpha )x-k\alpha \beta \gamma$

Comparing the coefficients of $x^3,x^2,x$ and constant terms on both sides, we get

$a=k,b=-k(\alpha +\beta +\gamma ),c=(\alpha \beta +\beta \gamma +\gamma \alpha )\text{ and }d=-k(\alpha \beta \gamma )$

$\Rightarrow \alpha +\beta +\gamma =-\frac{b}{a}\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$

And, $\alpha \beta \gamma =-\frac{d}{a}$

$\Rightarrow$ Sum of the zeros $=-\frac{b}{a}=-\frac{\text{ Coefficient of }{x}^2}{\text{ Coefficient of }{x}^3}$

$\Rightarrow$ Sum of the products of the zeros taken two at a time $=\frac{c}{a}=\frac{\text{ Coefficient of }x}{\text{ Coefficient of }{x}^3}$

$\Rightarrow$ Product of the zeros $=-\frac{d}{a}=-\frac{\text{ Constan }t\text{ term }}{\text{ Coefficient of }{x}^3}$

POLYNOMIALS

REMARKS :

Cubic polynomial having $\alpha ,\beta$ and $\gamma$ as its zeros is given by

$f(x)=k(x-\alpha )(x-\beta )(x-\gamma )$

$f(x)=k{x}^3-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +\gamma \alpha )x-\alpha \beta \gamma$ where $k$ is any non-zero real number.

POLYNOMIALS

Ex. 8 Verify that $\frac{1}{2},1-2$ are zeros of cubic polynomial $2x^3+x^2-5x+2$. Also verify the relationship between, the zeros and their coefficients.

POLYNOMIALS :

Sol. $f(x)=2x^3+x^2-5x+2$

$f(\frac{1}{2})=2(\frac{1}{2}{)}^3+(\frac{1}{2}{)}^2-5(\frac{1}{2})+2=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2$

$f(1)=2({)}^3+(1{)}^{2}5(1)+2=2+1-5+2=0$.

$f(-2)=2(-2{)}^3+(-2{)}^2-5(-2)+2=-16+4+10+2=0$.

Let $\alpha =\frac{1}{2},\beta =1$ and $\gamma =-2$

Now, Sum of zeros $\begin{array}{rl}& =\alpha +\beta +\gamma \\ & =\frac{1}{2}+1-2=-\frac{1}{2}\end{array}$

Also, sum of zeros

$=-\frac{(\text{ Coefficient of }{x}^2)}{\text{ Coefficient of }{x}^3}=-\frac{1}{2}$

So, sum of zeros

$=\alpha +\beta +\gamma =-\frac{(\text{ Coefficient of }{x}^2)}{\text{ Coefficient of }{x}^3}$

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Sum of product of zeros taken two at a time $=\alpha \beta +\beta \gamma +\gamma \alpha$

$=\frac{1}{2}\times 1+1\times (-2)+(-2)\times \frac{1}{2}=-\frac{5}{2}$

Also, $\beta \beta +\beta \gamma +\gamma \alpha =\frac{\text{ Coefficient of }x}{\text{ Coefficient of }{x}^3}=\frac{-5}{2}$

So, sum of product of zeros taken two at a time $=\alpha \beta +\beta \gamma +\gamma \alpha =\frac{\text{ Coefficient of }x}{\text{ Coefficient of }{x}^3}$

Now, Product of zeros $=\alpha \beta \gamma =(\frac{1}{2})(1)(-2)=-1$

Also, product of zeros $=\frac{\text{ Constan }t\text{ term }}{\text{ Coefficient of }{x}^3}=\frac{-2}{2}=-1$

$\therefore$ Product zeros $=\alpha \beta \gamma =-\frac{\text{ Constant term }}{\text{ Coefficient of }{x}^3}$

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Ex. 9 Find a polynomial with the sum, sum of the product of its zeros taken two at a time, and product its zeros as $3,-1$ and -3 respectively.

POLYNOMIALS

Sol. Given $\alpha +\beta +\gamma =3,\alpha \beta +\beta \gamma +\gamma \alpha =-1$ and $\alpha \beta \gamma =-3$

So, polynomial $f(x)=k{x}^3-x^2(\alpha +\beta +\gamma )+x(\alpha \beta +\beta \gamma +\gamma \alpha )-\alpha \beta \gamma$

$f(x)=k{x}^3-3x^2-x+3$, where $k$ is any non-zero real number.

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4.7 VALUE OF A POLYNOMIAL :

The value of a polynomial $\mathbf{f}(\mathbf{x})$ at $\mathbf{x}=\alpha$ is obtained by substituting $x=\alpha$ in the given polynomial and is denoted by $f(\alpha )$.

For example : If $f(x)=2x^3-13x^2+17x+12$ then its value at $x=1$ is

$f(1)=2(1{)}^3-13(1{)}^2+17(1)+12=2-13+17+12=18$.

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4.8 ZEROS OF ROOTS OF A POLYNOMIAL :

A real number ’ $\mathbf{a}$ ’ is a zero of a polynomial $f(x)$, if $f(a)=0$, Here ’ $a$ ’ is called a root of the equation $f(x)=0$.

Ex. 10 Show that $x=2$ is a root of $2x^3+x^2-7x-6$

Sol. $p(x)=2x^3+x^2-7x-6$.

Then, $p(2)=2(2{)}^3+(2{)}^2-7(2)-6=16+4-14=0$

Hence $x=2$ is a root of $p(x)$.

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Ex. 11 If $x=\frac{4}{3}$ is a root of the polynomial $f(x)=6x^3-11x^2+kx-20$ then find the value of $k$.

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Sol. $f(x)=6x^3-11x^2+kx-20$

$\begin{array}{rl}& f(\frac{4}{3})=6(\frac{4}{3}{)}^3-11(\frac{4}{3}{)}^2+k(\frac{4}{3})-20=0\\ & \Rightarrow 6(\frac{64}{27})-11(\frac{16}{9})+\frac{4k}{3}-20=0\Rightarrow 6(\frac{64}{27})-11(\frac{16}{9})+\frac{4k}{3}-20=0\\ & \Rightarrow 128-176+12k-180=0\Rightarrow 12k+128-356=0\\ & \Rightarrow 12k=228\Rightarrow k=19\end{array}$

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Ex. 12 If $x=2$ & $x=0$ are roots of the polynomials $(f)x=2x^3-5x^2+ax+b$, then find the values of $a$ and $b$ /

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Sol. $f(2)=2(2{)}^3-5(2{)}^2+a(2)+b=0$

$\Rightarrow 16-20+2a+b=0\Rightarrow 2a+b=4$ …(i)

$\Rightarrow f(0)=2(0{)}^3-5(0{)}^2+a(0)+b=0$

$\Rightarrow b=0$

$\Rightarrow 2a=4\Rightarrow a=2,b=0$.

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4.9 FACTOR THEOREM :

Let $\mathbf{p}(\mathbf{x})$ be a polynomial of degree greater than or equal to 1 and ’ $\mathbf{a}$ ’ be a real number such that $\mathbf{p}\mathbf{(}\mathbf{a}\mathbf{)}=\mathbf{0}$. then $(\mathbf{x}-\mathbf{a})$ is a factor of $\mathbf{p}(\mathbf{x})$. Conversely, if $(\mathbf{x}-\mathbf{a})$ is a factor of $\mathbf{p}(\mathbf{x})$, then $\mathbf{p}(\mathbf{a})\mathbf{=}\mathbf{0}$.

Ex. 13 Show that $x+1$ and $2x-3$ are factors of $2x^3-9x^2+x+12$.

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Sol. To prove that $(x+1)$ and $(2x-3)$ are factors of $p(x)=2x^3-9x^2+x+12$ it is sufficient to show that $p(-1)$ and $p(\frac{3}{2})$ both are equal to zero.

$p(-1)=2(-1{)}^3-9(-1{)}^2+(-1)+12=-2-9-1+12=-12+12=0$

And $p(\frac{3}{2})=2(\frac{3}{2}{)}^3-9(\frac{3}{2}{)}^2+(\frac{3}{2})+12$

$=\frac{27}{4}-\frac{81}{4}+\frac{3}{2}+12=\frac{27-81+6+48}{4}=\frac{-81+81}{4}=0$

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Ex. 14 Find $\alpha$ and $\beta$ if $x+1$ and $x+2$ are factors of $p(x)=x^3+3x^2-2\alpha x+\beta$.

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Sol. $x+1$ and $x+2$ are the factor of $p(x)$.

Then, $p(-1)=0$ & $p(-2)=0$

Therefore, $p(-1)=(-1{)}^3+3(-1{)}^2-2\alpha (-1)+\beta =0$

$\Rightarrow -1+3+2\alpha +\beta =0\Rightarrow \beta =-2\alpha -2$ …(i)

$p(-2)=(-2{)}^3+3(-2{)}^2-2\alpha (-2)+\beta =0$

$\Rightarrow -8+12+4\alpha +\beta =0\Rightarrow \beta =-4\alpha -4$ …(ii)

From equation (1) and (2)

$-2\alpha -2=-4\alpha -4\Rightarrow 2\alpha =-2\Rightarrow \alpha =-1$

Put $\alpha =-1$ equation (1) $\Rightarrow \beta =-2(-1)-2=2-2=0$. Hence $\alpha =-1,\beta =0$

POLYNOMIALS

Ex. 15 What must be added to $3x^3+x^2-22x+9$ so that the result is exactly divisible by $3x^2+7x-6$.

POLYNOMIALS

Sol. Let $p(x)=3x^3+x^2-22x+9$ and $q(x)=3x^2+7x-6$

We know if $p(x)$ is divided by $q(x)$ which is quadratic polynomial then the remainder be $r(x)$ and degree of $r(x)$ is less than $q(x)$ or Divisor.

$\therefore$ By long division method

Let we added $ax+b$ (linear polynomial) in $p(x)$, so that $p(x)+ax+b$ is exactly divisible by $3x^2+7x-6$.

Hence, $p(x)+ax+b=s(x)=3x^3-x^2-22x+9+ax+b=3x^3+x^{2}x(22-a)+(9+b)$.

POLYNOMIALS

Hence, $x(a-2)+b-3=0.x+0$

$\Rightarrow a-2=0$ & $b-3=0\Rightarrow a=2$ and $b=3$

Hence if in $p(x)$ we added $2x+3$ then it is exactly divisible by $3x^2+7x-6$.

POLYNOMIALS

Ex. 16 What must be subtracted from $x^3-6x^2-15x+80$ so that the result is exactly divisible by $2+x-12$.

POLYNOMIALS

Sol. Let $ax+b$ be subtracted from $p(x)=x^3-6x^2-15x+80$ so that it is exactly divisible by $x^2+x-12$.

$\therefore s(x)=x^3-6x^2-15x+80-(ax+b)$

$=x^3-6x^2-(15+a)x+(80-b)$

Dividend $=$ Divisor $\times$ quotient + remainder

But remainder will be zero.

$\therefore$ Dividend $=$ Divisor $\times$ quotient

$\Rightarrow s(x)=(x^2+x-12)\times \text{ quotient }\Rightarrow s(x)=x^3-6x^2-(15+a)x+(80-b)$

POLYNOMIALS

Hence, $x(4-a)+(-4-b)=0.x+0$

$\Rightarrow 4-a=0$ & $(-4-b)=0\Rightarrow a=4$ and $b=-4$

Hence, if in $p(x)$ we subtract $4x-4$ then it is exactly divisible by $x^2+x-12$.

POLYNOMIALS

Ex. 17 Using factor theorem, factorize : $p(x)=2x^4-7x^3-13x^2+63x-45$.

POLYNOMIALS

Sol. $45\Rightarrow \pm 1,\pm 3,\pm 5,\pm 9,\pm 15,\pm 45$

If we put $x=1$ in $p(x)$

$p(1)=2(1{)}^4-7(1{)}^3-13(1{)}^2+63(1)-45$

$p(1)=2-7-13+63-45=65-65=0$

$\therefore x=1$ or $x-1$ is a factor of $p(x)$.

Similarly if we put $x=3$ in $p(x)$

$p(3)=2(3{)}^4-7(3{)}^3-13(3{)}^2+63(3)-45$

$p(3)=162-189-117+189-45=162-162=0$

POLYNOMIALS

Hence, $x=3$ or $(x-3)=0$ is the factor of $p(x)$.

$p(x)=2x^4-7x^3-13x^2+63x-45$

$\therefore p(x)=2x^3(x-1)-5x^2(x-1)-18x(x-1)+45(x-1)$

$\Rightarrow p(x)=(x-1)(2x^3-5x^2-18x+45)$ $\Rightarrow p(x)=(x-1)(2x^3-5x^2-18x+45)$

$\Rightarrow p(x)=(x-1)[2x^2-(x-3)+x(x-3)-15(x-3)]$ $\Rightarrow p(x)=(x-1)(x-3)(2x^2+x-15)$

$\Rightarrow p(x)=(x-1)(x-3)(2x^2+6x-5x-15)$ $\Rightarrow p(x)=(x-1)(x-3)[2x(x+3)-5(x+3)]$

$\Rightarrow p(x)=(x-1)(x-3)(x+3)(2x-5)$.

POLYNOMIALS

4.10 REMAINDER THEOREM :

Let $\mathbf{p}(\mathbf{x})$ be any polynomial of degree greater than or equal to one and ’ $\mathbf{a}$ ’ be any real number. If $\mathbf{p}(\mathbf{x})$ is divided by $\mathbf{x}-\mathbf{a})$, then the remainder is equal to $\mathbf{p}\mathbf{(}\mathbf{a}\mathbf{)}$.

Let $q(x)$ be the quotient and $r(X)$ be the remainder when $p(x)$ is divided by $(x-a)$, then

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

POLYNOMIALS

Ex. 18 Find the remainder when $f(x)=x^3-6x^2+2x-4$ is divided by $g(x)=1-2x$.

POLYNOMIALS

Sol. $1-2x=0\Rightarrow 2x=1\Rightarrow x=\frac{1}{2}$

$f(\frac{1}{2})=(\frac{1}{2}{)}^3-6(\frac{1}{2}{)}^2+2(\frac{1}{2})-4=\frac{1}{8}-\frac{3}{2}+1-4=\frac{1-12+8-32}{8}=-\frac{35}{8}$

POLYNOMIALS

Ex. 19 Apply division algorithm to find the quotient $q(x)$ and remainder $r(x)$ on dividing $f(x)=10x^4+17x^3$. $62x^2+30x-3$ by $b(x)=2x^2-x+1$.