Solution

(A) $\dfrac{x^{2}}{2}-\dfrac{2}{x^{2}}$ can be written as $\dfrac{x^{2}}{2}-2 x^{-2}$. The power of $x^{-2}$ is -2 , which is not a whole number. So, the given algebraic expression is not a polynomial.

(B) $\sqrt{2 x}-1$ can be written as $\sqrt{2} x^{\dfrac{1}{2}}-1$. The power of $x^{\dfrac{1}{2}}$ is $\dfrac{1}{2}$, which is not a whole number. So, the given algebraic expression is not a polynomial.

(C) $x^{2}+\dfrac{3 x^{\dfrac{3}{2}}}{\sqrt{x}}$ can be written as $x^{2}+3 x$. The power of the variable are whole number. So, the given algebraic expression is a polynomial.

(D) $\dfrac{x-1}{x+1}$ is not a standard from of a polynomial.

Hence, the correct option is (D).

Solution

The term $\sqrt{2}$ can be written as $\sqrt{2} x^{0}$. Since, the power of $x$ is 0 . Therefore, the degree of the constant term polynomial is 0 .

Hence, the correct option is (D).

Solution

We know that, the degree of given polynomial $4 x^{4}+0 \times x^{3}+0 \times x^{5}+5 x+7$ will be the highest power of variable that is 4 .

Hence, the correct option is (A).

Solution

The degree of the polynomial is equal to the highest power of the variable. We know that, the degree of zero polynomial is not defined.

Hence, the correct option is (D).

Solution

Consider the polynomial:

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

The value of $p(2 \sqrt{2})$ is calculated as follows:

$ \begin{aligned} p(2 \sqrt{2}) & =(2 \sqrt{2})^{2}-2 \sqrt{2} \times 2 \sqrt{2}+1 \\ & =8-8+1 \\ & =1 \end{aligned} $

Hence, the correct option is (B).

Solution

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

The value of $f(-1)$ will be:

$ \begin{aligned} f(-1) & =5 \times(-1)-4 \times(-1)^{2}+3 \\ & =-5-4+3 \\ & =-9+3 \\ & =6 \end{aligned} $

Hence, the correct option is (B).

Solution

Given:

$p(x)=x+3$

So,

$p(-x)=(-x)+3$

$p(-x)=-x+3$

Now, the sum of $p(x)+p(-x)$ is:

$ \begin{aligned} p(x)+p(-x) & =x+3+(-x+3) \\ & =x+3-x+3 \\ & =6 \end{aligned} $

Hence, the correct option is (D).

Solution

The zero degree of the zero polynomial is not defined.

Solution

Zero of the polynomial $p(x)=2 x+5$ is $p(x)=0$. So,

$2 x+5=0$

$2 x=-5$

$x=-\dfrac{5}{2}$

Hence, the correct option is (B).

Solution

Let $f(x)=2 x^{2}+7 x-4$

Zero of the polynomial is calculate as $f(x)=0$.

$ \begin{aligned} f(x) & =0 \\ 2 x^{2}+7 x-4 & =0 \\ 2 x^{2}+8 x-x-4 & =0 \\ 2 x(x+4)-1(x+4) & =0 \\ (2 x-1)(x+4) & =0 \\ 2 x-1 & =0 \text{ or } x+4=0 \\ 2 x & =1 \text{ or } x=-4 \\ x & =\dfrac{1}{2} \text{ or } x=-4 \end{aligned} $

Hence, the correct option is (B).

Solution

Let $f(x)=x^{51}+51$

We know that when $f(x)$ is divided by $x+a$, then the remainder is $f(-a)$.

So, remainder is:

$ \begin{aligned} f(-1) & =(-1)^{51}+51 \\ & =-1+51 \\ & =50 \end{aligned} $

Hence, the correct option is (D).

Solution

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

Zero of $x+1$ is $x=-1$.

By factor theorem $f(-1)=0$. So,

$ \begin{aligned} f(x) & =0 \\ 2 \times(-1)^{2}+k \times(-1) & =0 \\ 2-k & =0 \\ 2 & =k \end{aligned} $

Hence, the correct option is (C).

Solution

We know that if $x+a$ is a factor of $f(x)$ then, $f(-a)=0$.

(A) Let $f(x)=x^{3}+x^{2}-x+1$

Now,

$ \begin{aligned} f(-1) & =(-1)^{3}+(-1)^{2}-(-1)+1 \\ & =-1+1+1+1 \\ & =2 \neq 0 \end{aligned} $

So, $f(x)$ is not a factor of $x+1$.

(B) Let $f(x)=x^{3}+x^{2}+x+1$

Now,

$ \begin{aligned} f(-1) & =(-1)^{3}+(-1)^{2}+(-1)+1 \\ & =-1+1-1+1 \\ & =0 \end{aligned} $

So, $f(x)$ is a factor of $x+1$.

(C) Let $f(x)=x^{4}+x^{3}+x^{2}+1$

Now,

$ \begin{aligned} f(-1) & =(-1)^{4}+(-1)^{3}+(-1)^{2}+1 \\ & =1-1+1+1 \\ & =2 \neq 0 \end{aligned} $

So, $f(x)$ is not a factor of $x+1$.

(D) Let $f(x)=x^{4}+3 x^{3}+3 x^{2}+x+1$

Now,

$ \begin{aligned} f(-1) & =(-1)^{4}+3 \times(-1)^{3}+3 \times(-1)^{2}+(-1)+1 \\ & =1-3+3-1+1 \\ & =1 \neq 0 \end{aligned} $

So, $f(x)$ is not a factor of $x+1$.

Hence, the correct option is (B).

Solution

$ \begin{aligned} (25 x^{2}-1)+(1+5 x)^{2} & =(5 x)^{2}-1^{2}+(5 x+1)^{2} \\ & =(5 x-1)(5 x-1)+(5 x+1)^{2} \\ & =(5 x+1)(5 x-1+5 x+1) \\ & =(5 x+1) 10 x \\ & =10 x(5 x+1) \end{aligned} $

So, one of the factor of $(25 x^{2}-1)+(1+5 x)^{2}$ is $10 x$.

Hence, the correct option is (D).

Solution

Consider the expression:

$249^{2}-248^{2}$

Use the identity: $a^{2}-b^{2}=(a+b)(a-b)$

$ \begin{aligned} 249^{2}-248^{2} & =(249+248)(249-248) \\ & =497 \times 1 \\ & =497 \end{aligned} $

Hence, the correct option is (D).

Solution

Do the factor of given expression as follows:

$ \begin{aligned} 4 x^{2}+8 x+3 & =4 x^{2}+6 x+2 x+3 \\ & =2 x(2 x+3)+1(2 x+3) \\ & =(2 x+1)(2 x+3) \end{aligned} $

Hence, the correct option is (B).

Solution

$ \begin{aligned} (x+y)^{3}-(x^{3}+y^{3}) & =x^{3}+y^{3}+3 x y(x+y)-x^{3}-y^{3} \\ & =3 x y(x+y) \end{aligned} $

Since, $3 x y$ is factor of $(x+y)^{3}-(x^{3}+y^{3})$.

Hence, the correct option is (D).

Solution

Consider the expression:

$(x+3)^{3}$

Use the identity: $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$

$(x+3)^{3}=x^{3}+3^{3}+3 \times x \times 3(x+3)$

$ =x^{3}+27+9 x^{2}+27 x $

Since, the coefficient of $x$ is 27 .

Hence, the correct option is (D).

Solution

Consider the equation:

$\dfrac{x}{y}+\dfrac{y}{x}=-1$

Simplify the above expression as follows:

$ \dfrac{x^{2}+y^{2}}{x y}=-1 $

$ x^{2}+y^{2}=-x y $

Now,

$ \begin{aligned} x^{3}-y^{3} & =(x-y)(x^{2}+y^{2}+x y) \\ & =(x-y)(-x y+x y)[\text{ Substitute: } x^{2}+y^{2}=-x y] \\ & =(x-y) \times 0 \\ & =0 \end{aligned} $

Hence, the correct option is (C).

Solution

Consider the equation:

$49 x^{2}-b=(7 x+\dfrac{1}{2})(7 x-\dfrac{1}{2})$

Solve the above equation as follows:

$49 x^{2}-b=(7 x+\dfrac{1}{2})(7 x-\dfrac{1}{2})[.$ Use the identity: $.(a+b)(a-b)=a^{2}-b^{2}]$

$49 x^{2}-b=(7 x)^{2}-(\dfrac{1}{2})^{2}$

$49 x^{2}-b=49 x^{2}-\dfrac{1}{4}$

Compare the each term of both side, get:

$b=\dfrac{1}{4}$

Hence, the correct option is (C).

Solution

We know that:

$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-c a)$

As $a+b+c=0$, So, $a^{3}+b^{3}+c^{3}-3 a b c=(0)(a^{2}+b^{2}+c^{2}-a b-b c-c a)=0$

Hence, $a^{3}+b^{3}+c^{3}=3 a b c$

Therefore, the correct option is (C).

Solution

(i)

Consider the expression:

8

The expression 8 also can be written as $8 x^{0}$.

Since, the power of $x$ is 0 , which is a whole number.

Therefore, the expression 8 is a polynomial.

(ii)

Consider the expression:

$\sqrt{3} x^{2}-2 x$

In the above expression, the power of $x$ are 2 and 1 respectively, and they are whole number. Therefore, the expression $\sqrt{3} x^{2}-2 x$ is a polynomial.

(iii)

Consider the expression:

$1-\sqrt{5 x}$

$1-\sqrt{5 x}$ can be written as $1-\sqrt{5} x^{\dfrac{1}{2}}$.

Since, the power of $x$ is $1 / 2$, which is not a whole number.

Therefore, the expression $1-\sqrt{5} x$ is not a polynomial.

(iv)

Consider the expression:

$\dfrac{1}{5 x^{-2}}+5 x+7$

$\dfrac{1}{5 x^{-2}}+5 x+7$ can be written as $5 x^{2}+5 x+7$.

Since, the power of $x$ are 2 and 1 respectively, and they are whole number.

Therefore, the expression $\dfrac{1}{5 x^{-2}}+5 x+7$ is a polynomial.

(v)

Consider the expression:

$\dfrac{(x-2)(x-4)}{x}$

Simplify the above expression as follows:

$ \begin{aligned} \dfrac{(x-2)(x-4)}{x} & =\dfrac{x^{2}-4 x-2 x+8}{x} \\ & =\dfrac{x^{2}-6 x+8}{x} \\ & =x-6+8 x^{-1} \end{aligned} $

Since, the power of $x=-1$ is not a whole number.

Therefore, the expression $\dfrac{(x-2)(x-4)}{x}$ is not a polynomial.

(vi)

Consider the expression:

$\dfrac{1}{1+x}$

Simplify the above expression as follows:

$\dfrac{1}{1+x}=(1+x)^{-1}$

Since, the power $x$ is not a whole number.

Therefore, the expression $\dfrac{1}{1+x}$ is not a polynomial.

(vii)

Consider the expression:

$\dfrac{1}{7} a^{3}-\dfrac{2}{\sqrt{3}} a^{2}+4 a-7$

In the above expression, the power of a are 3,2 , and 1 , which are whole number.

Therefore, the expression $\dfrac{1}{7} a^{3}-\dfrac{2}{\sqrt{3}} a^{2}+4 a-7$ is a polynomial.

(viii)

Consider the expression: $\dfrac{1}{2 x}$

The expression $\dfrac{1}{2 x}$ can be written as $\dfrac{1}{2 x}=\dfrac{1}{2} x^{-1}$.

Since, the power of $x$ is -1 , which is not a whole number.

Therefore, the expression $\dfrac{1}{2 x}$ is not a polynomial.

Solution-

(i) We know that the binomial have only two term so, the given statement is false. For example: $(a+b)^{n}$.

(ii) A polynomial can be a power of $0,1,2$, and 3 etc. that are also called respectively monomial, binomial, and trinomial. For example $x^{3}+x^{2}+x+1$ is a polynomial but not binomial. Therefore, the given statement is false.

(iii) We know that binomial have two term whose degree is a whole number greater than equal to 1 . For example, $x^{5}-1$ is a binomial of degree 5 . Hence, the given statement is true.

(iv) The zero of polynomial can be any real number. Since, the given statement is false.

(v) A polynomial can have any number of zero and it’s dependent on the degree of polynomial. Hence, the given statement is false.

(vi) The given statement is false because it will be not always true. For example: $x^{5}+2 x^{4}+3 x^{3}+x^{2}+x+1$ and $-x^{5}-2 x^{4}+x^{3}+x^{2}+x+1$, the degree of both polynomial is 5 but their sum is $4 x^{3}+2 x^{2}+2 x+2$. Therefore, the degree of this polynomial is 3 not 5 .

Solution

(i)

The given polynomial $x^{2}+x+1$ has only one variable that is $x$.

Therefore, the given polynomial is a polynomial in one variable.

(ii)

The given polynomial $y^{3}-5 y$ has only one variable that is $y$.

Therefore, the given polynomial is a polynomial in one variable.

(iii)

The given polynomial $x y+y z+z x$ has two variable that are $x$ and $y$.

Therefore, the given polynomial is a polynomial in two variable.

(iv)

The given polynomial $x^{2}-2 x y+y^{2}+1$ has two variable that are $x$ and $y$.

Therefore, the given polynomial is a polynomial in two variable.

Solution

(i)

Consider the expression: $2 x-1$

The degree of a polynomial in one variable is equal to highest power of the variable in algebraic expression that is 1 .

(ii)

Consider the expression: -10

-10 can be written as $-10 x^{0}$.

The degree of a polynomial in one variable is equal to highest power of the variable in algebraic expression that is 0 .

(iii)

Consider the expression: $x^{3}-9 x+3 x^{5}$

The degree of a polynomial in one variable is equal to highest power of the variable in algebraic expression that is 5 .

(iv)

Consider the expression: $y^{3}(1-y^{4})$

Simplify the above expression as follows:

$y^{3}(1-y^{4})=y^{3}-y^{7}$

The degree of a polynomial in one variable is equal to highest power of the variable in algebraic expression that is 7 .

Solution

Consider the expression:

$\dfrac{x^{3}+2 x+1}{5}-\dfrac{7}{2} x^{2}-x^{6}$

Simplify the above expression as:

$ \begin{aligned} \dfrac{x^{3}+2 x+1}{5}-\dfrac{7}{2} x^{2}-x^{6} & =\dfrac{1}{5} x^{3}+\dfrac{2}{5} x+\dfrac{1}{5}-\dfrac{7}{2} x^{2}-x^{6} \\ & =-x^{6}+\dfrac{1}{5} x^{3}-\dfrac{7}{2} x^{2}+\dfrac{2}{5} x+\dfrac{1}{5} \end{aligned} $

(i)

The degree of a polynomial in one variable is equal to highest power of the variable in algebraic expression that is 6 .

(ii)

The coefficient of $x^{3}$ in the given expression is $\dfrac{1}{5}$.

The coefficient of $x^{3}$ in the given expression is -1 . (iv)

The constant term in the given expression is $\dfrac{1}{5}$ because it has no variable $x$ associated with it.

Solution

(i)

In the given polynomial $\dfrac{\pi}{6} x+x^{2}-1$, the coefficient of $x^{2}$ is 1 .

(ii)

The polynomial $3 x-5$ can be written as $0 x^{2}+3 x-5$.

Therefore, the coefficient of $x^{2}$ is 0 .

(iii)

Consider the expression:

$(x-1)(3 x-4)$

Simplify the above expression as:

$(x-1)(3 x-4)=3 x^{2}-4 x-3 x+4$

$=3 x^{2}-7 x+4$

Therefore, the coefficient of $x^{2}$ is 3 .

(iv)

Consider the expression:

$(2 x-5)(2 x^{2}-3 x+1)$

Simplify the above expression as:

$ \begin{aligned} (2 x-5)(2 x^{2}-3 x+1) & =4 x^{3}-6 x^{2}+2 x-10 x^{2}+15 x-5 \\ & =4 x^{3}-16 x^{2}+17 x-5 \end{aligned} $

Therefore, the coefficient of $x^{2}$ is -16 .

Solution

(i)

Consider the polynomial - $2-x^{2}+x^{3}$

The highest power in the above polynomial is 3 . So, the given polynomial is cubic polynomials.

(ii)

Consider the polynomial $-3 x^{3}$

The highest power in the above polynomial is 3 . So, the given polynomial is cubic polynomials.

(iii)

Consider the polynomial $-5 t-\sqrt{7}$

The highest power in the above polynomial is 1 . So, the given polynomial is linear polynomials.

(iv)

Consider the polynomial - $4-5 y^{2}$

The highest power in the above polynomial is 2 . So, the given polynomial is quadratic polynomials.

(v)

Consider the polynomial - $\mathbf{3}$

3 can be written as $3 x^{0}$.

The highest power in the above polynomial is 0 . So, the given polynomial is constant polynomials.

(vi)

Consider the polynomial - $\mathbf{2}+\mathbf{x}$

The degree of the above polynomial is 1 . So, the given polynomial is linear polynomials.

(vii)

Consider the polynomial $-y^{3}-y$

The degree of the above polynomial is 3 . So, the given polynomial is cubic polynomials.

(viii)

Consider the polynomial $-1+x+x^{3}$

The degree of the above polynomial is 3 . So, the given polynomial is cubic polynomials.

(ix)

Consider the polynomial - $t^{3}$

The degree of the above polynomial is 3 . So, the given polynomial is cubic polynomials.

(x)

Consider the polynomial $-\sqrt{2} x-1$

The degree of the above polynomial is 1 . So, the given polynomial is linear polynomials.

Solution

(i)

We know that monomial polynomial has only one term. For example of monomial polynomial with degree 1 is $2 x$.

(ii)

We know that binomial polynomial has only two term. For example of binomial polynomial with degree 20 is $x^{20}+10$.

(iii)

We know that Trinomial polynomial has only two term. For example of Trinomial polynomial with degree 2 is $x^{2}+2 x+5$.

Solution

Consider the polynomial:

$p(x)=3 x^{3}-4 x^{2}+7 x-5$

The value of given polynomial at $x=3$ is:

$ \begin{aligned} p(3) & =3 \times 3^{3}-4 \times 3^{2}+7 \times 3-5 \\ & =3 \times 27-4 \times 9+7 \times 3-5 \\ & =61 \end{aligned} $

Now, the value of given polynomial at $x=-3$ is:

$ \begin{aligned} p(-3) & =3 \times(-3)^{3}-4 \times(-3)^{2}+7 \times(-3)-5 \\ & =3 \times(-27)-4 \times 9+7 \times(-3)-5 \\ & =-143 \end{aligned} $

Solution

Consider the polynomial:

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

The value of given polynomial at $x=2$ is:

$ \begin{aligned} p(2) & =2^{2}-4 \times 2+3 \\ & =4-8+3 \\ & =-1 \end{aligned} $

When $x=-1$,

$ \begin{aligned} p(-1) & =(-1)^{2}-4 \times(-1)+3 \\ & =1+4+3 \\ & =8 \end{aligned} $

When $x=\dfrac{1}{2}$,

$ \begin{aligned} p(-1) & =(\dfrac{1}{2})^{2}-4 \times(\dfrac{1}{2})+3 \\ & =\dfrac{1}{4}-2+3 \\ & =\dfrac{5}{4} \end{aligned} $

Now, the value of $p(2)-p(-1)+p(\dfrac{1}{2})$ will be:

$ \begin{aligned} p(2)-p(-1)+p(\dfrac{1}{2}) & =-1-8+\dfrac{5}{4} \\ & =-9+\dfrac{5}{4} \\ & =\dfrac{-36+5}{4} \\ & =-\dfrac{31}{4} \end{aligned} $

Solution

(i)

Consider the polynomial:

$p(x)=10 x-4 x^{2}-3$

The value of given polynomial at $x=0$ is:

$ \begin{aligned} p(0) & =10 \times 0-4 \times 0^{2}-3 \\ & =-3 \end{aligned} $

When $x=1$

$ \begin{aligned} p(1) & =10 \times 1-4 \times 1^{2}-3 \\ & =10-4-3 \\ & =3 \end{aligned} $

When $x=-2$

$ \begin{aligned} p(-2) & =10 \times(-2)-4 \times(-2)^{2}-3 \\ & =-20-16-3 \\ & =-39 \end{aligned} $

(ii)

Consider the polynomial:

$ p(y)=(y+2)(y-2) $

The value of given polynomial at $y=0$ is:

$ \begin{aligned} p(0) & =(0+2)(0-2) \\ & =2 \times-2 \\ & =-4 \end{aligned} $

$ \begin{aligned} & \text{ When } y=1 \\ & \begin{aligned} p(1)= & (1+2)(1-2) \\ & =3 \times-1 \\ & =3 \end{aligned} \\ & \begin{aligned} & \text{ When } y=-2 \\ & p(1)=(-2+2)(-2-2) \\ &=0 \times(-2-2) \\ &=0 \end{aligned} \end{aligned} $

Solution

(i)

Zero of $x-3$ is calculated as follows:

$x-3=0$

$x=3$

Hence, the given statement is false.

(ii)

Zero of $3 x+1$ is calculated as follows:

$3 x+1=0$

$3 x=-1$

$x=-\dfrac{1}{3}$

Hence, the given statement is true.

(iii)

Zero of $4-5 y$ is calculated as follows:

$4-5 y=0$

$4=5 y$

$y=\dfrac{4}{5}$

Hence, the given statement is false.

(iv)

Zero of $t^{2}-2 t$ is calculated as follows: $t^{2}-2 t=0$

$t(t-2)=0$

$t=0$ or $t-2=0$

$t=0$ or $t=2$

Hence, the given statement is true.

(v)

Zero of $y^{2}+y-6$ is calculated as follows:

$ \begin{aligned} y^{2}+y-6 & =0 \\ y^{2}+3 y-2 y-6 & =0 \\ y(y+3)-2(y+3) & =0 \\ (y+3)(y-2) & =0 \\ y+3 & =0 \text{ or } y-2=0 \\ y & =-3 \text{ or } y=2 \end{aligned} $

Hence, the given statement is true.

Solution

(i)

Consider the polynomial: $p(x)=x-4$

The zeroes of the given polynomial $p(x)=x-4$ is calculated as follows:

$p(x)=0$

$x-4=0$

$x=4$

(ii)

Consider the polynomial: $g(x)=3-6 x$

The zeroes of the given polynomial $g(x)=3-6 x$ is calculated as follows: $g(x)=0$

$3-6 x=0$

$3=6 x$

$x=\dfrac{3}{6}$

$x=\dfrac{1}{2}$

(iii)

Consider the polynomial: $q(x)=2 x-7$

The zeroes of the given polynomial $q(x)=2 x-7$ is calculated as follows:

$ \begin{aligned} q(x) & =0 \\ 2 x-7 & =0 \\ 2 x & =7 \\ x & =\dfrac{7}{2} \end{aligned} $

(iv)

Consider the polynomial:

$h(y)=2 y$

The zeroes of the given polynomial $h(y)=2 y$ is calculated as follows:

$ \begin{aligned} h(y) & =0 \\ 2 y & =0 \\ y & =0 \end{aligned} $

Solution

Consider the polynomial:

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

The zeroes of the polynomial can be calculated as:

$ p(x)=0 $

$(x-2)^{2}-(x+2)^{2}=0$

Used the identity: $a^{2}-b^{2}=(a+b)(a-b)$

Solution

Let $f(z)=a z^{3}+4 z^{2}+3 z-4$ and $g(z)=z^{3}-4 z+a$

Both given polynomial have leave same remainder when divided by $z-3$. So, $f(z)=$ $g(z)$.

$ \begin{aligned} f(3) & =a(3)^{3}+4 \times(3)^{2}+3 \times 3-4 \\ & =27 a+36+9-4 \\ & =27 a+41 \end{aligned} $

And:

$ \begin{aligned} q(3) & =(3)^{3}-4 \times 3+a \\ & =27-12+a \\ & =15+a \end{aligned} $

Now, $f(3)=g(3)$

$27 a+41=15+a$

$ \begin{aligned} 26 a & =-26 \\ a & =-1 \end{aligned} $

Therefore, the required value of $a=-1$.

Solution

We know that when $p(x)$ is divided by $x+a$, then the remainder $p(-a)$.

When given polynomial $p(x)=x^{4}-2 x^{3}+3 x^{2}-a x+3 a-7$ is divided by $x+1$, then the remainder $=p(-1)$

Now,

$ \begin{aligned} p(-1) & =(-1)^{4}-2 \times(-1)^{3}+3 \times(-1)^{2}-a \times(-1)+3 a-7 \\ & =1-2 \times(-1)+3 \times(1)+a+3 a-7 \\ & =1+2+3+4 a-7 \\ & =-1+4 a \end{aligned} $

Also, remainder $=19$

$ \begin{aligned} -1+4 a & =19 \\ 4 a & =20 \\ a & =\dfrac{20}{4} \\ a & =5 \end{aligned} $

Now, when $p(x)$ is divided by $x+2$, then:

$ \begin{aligned} & \text{ Remainder }=p(-2) \\ & =(-2)^{4}-2 \times(-2)^{3}+3 \times(-2)^{2}-a \times(-2)+3 a-7 \\ & =16+16+12+2 a+3 a-7 \\ & =37+5 a \\ & =37+5 \times 5 \\ & =37+25 \\ & =62 \end{aligned} $

Solution

Let $p(x)=p x^{2}+5 x+r$

Given $(x-2)$ is a factor of $p(x)$.

So, $p(2)=0$

$p(2)^{2}+5 \times 2+r=0$

$4 p+10+r=0 \ldots(1)$

Again, $(x-\dfrac{1}{2})$ is a factor of $p(x)$.

So, $p(\dfrac{1}{2})=0$

Now,

$ \begin{gathered} p(\dfrac{1}{2})=p(\dfrac{1}{2})^{2}+5 \times(\dfrac{1}{2})+r \\ =\dfrac{1}{4} p+\dfrac{5}{2}+r \\ p(\dfrac{1}{2})=0 \\ \dfrac{1}{4} p+\dfrac{5}{2}+r=0 \end{gathered} $

From (1), we have $4 p+r=-10$

From (2), we have $p+10+4 r=0$ $p+4 r=-10$

$4 p+r=p+4 r$

$3 p=3 r$

$p=r$

Hence, proved.

Solution

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

Consider the polynomial:

$x^{2}-3 x+2$

Then, factorise the above expression as:

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

We know that when $p(x)$ is divided by $x+a$, then the remainder $=p(-a)$.

Now, $p(1)=2 \times 1^{4}-5 \times 1^{3}+2 \times 1^{2}-1+2$

$ =2-5+2-1+2 $

$ =0 $

$P(1)=0$

Therefore, $(x-1)$ divides $p(x)$. So,

$(x-1)(x-2)=x^{2}-3 x+2$ Divides $2 x^{4}-5 x^{3}+2 x^{2}-x+2$

Solution

Consider the expression:

$ (2 x-5 y)^{3}-(2 x+5 y)^{3} $

Use the identity: $a^{3}-b^{3}=(a-b)(a^{2}+a b+b^{2})$

$ \begin{aligned} & (2 x-5 y)^{3}-(2 x+5 y)^{3}={(2 x-5 y)-(2 x+5 y)}{(2 x-5 y)^{2}+(2 x-5 y)(2 x+5 y)+(2 x+5 y)^{2}} \\ & =(2 x-5 y-2 x-5 y)(4 x^{2}+25 y^{2}-20 x y+4 x^{2}-25 y^{2}+4 x^{2}+25 y^{2}+20 x y) \\ & =(-10 y)(2 x^{2}+25 y^{2}) \\ & =-120 x^{2} y-250 y^{3} \end{aligned} $

Solution

According to the question:

$(x^{2}+4 y^{2}+z^{2}+2 x y+x z-2 y z) \times(-z+x-2 y)$

Now, multiply as follows:

$={x+(-2 y)+(-z)}{(x)^{2}+(-2 y)^{2}+(-z^{2})-(x)(-2 y)-(-2 y)(-z)-(-z)(x)}$

$=x^{3}+(-2 y)^{3}+(-z)^{3}-3 \times x \times(-2 y) \times(-z)$

Use the identity: $(a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-c a)=a^{3}+b^{3}+c^{3}-3 a b c$

$=x^{3}-8 y^{3}-z^{3}-6 x y z$

Solution

Given in the question, $a, b$ and $c$ is non-zero and $a+b+c=0$.

Therefore, $a^{3}+b^{3}+c^{3}=3 a b c$

Now,

$ \begin{aligned} \dfrac{a^{2}}{b c}+\dfrac{b^{2}}{c a}+\dfrac{c^{2}}{a b} & =\dfrac{a^{2}+b^{2}+c^{2}}{a b c} \\ & =\dfrac{3 a b c}{a b c} \\ & =3 \end{aligned} $

Solution

Given:

$a+b+c=5$ and $ab+bc+ca=10$

We know that:

$ \begin{aligned} a^{3}+b^{3}+c^{3}-3 a b c & =(a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-c a) \\ & =(a+b+c)[a^{2}+b^{2}+c^{2}-(a b+b c+c a)] \\ & =5{a^{2}+b^{2}+c^{2}-(a b+b c+c a)} \\ & =5(a^{2}+b^{2}+c^{2}-10) \end{aligned} $

Given: $a+b+c=5$

Now, squaring both sides, get:

$ \begin{aligned} (a+b+c)^{2} & =5^{2} \\ a^{2}+b^{2}+c^{2}+2(a b+b c+c a) & =25 \\ a^{2}+b^{2}+c^{2}+2 \times 10 & =25 \\ a^{2}+b^{2}+c^{2} & =25-20 \\ & =5 \end{aligned} $

Now,

$ \begin{aligned} a^{3}+b^{3}+c^{3}-3 a b c & =5(a^{2}+b^{2}+c^{2}-10) \\ & =5 \times(5-10) \\ & =5 \times(-5) \\ & =-25 \end{aligned} $

Hence, proved.

Solution

$ \begin{aligned} (a+b+c)^{3} & =[a+(b+c)]^{3} \\ & =a^{3}+3 a^{2}(b+c)+3 a(b+c)^{2}+(b+c)^{3} \\ & =a^{3}+3 a^{2} b+3 a^{2} c+3 a(b^{2}+2 b c+c^{2})+(b^{3}+3 b^{2} c+3 b c^{2}+c^{3}) \\ & =a^{3}+3 a^{2} b+3 a^{2} c+3 a b^{2}+6 a b c+3 a c^{2}+b^{3}+3 b^{2} c+3 b c^{2}+c^{3} \\ & =a^{3}+b^{3}+c^{3}+3 a^{2} b+3 a^{2} c+3 b^{2} c+3 b^{2} a+3 c^{2} a+3 c^{2} b+6 a b c \\ & =a^{3}+b^{3}+c^{3}+3 a^{2}(b+c)+3 b^{2}(c+a)+3 c^{2}(a+b)+6 a b c \end{aligned} $

Since, the above result can be written as:

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

Therefore, $(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(b+c)(c+a)$