$p(x)=x^{3}+2 x^{2}+k x+3$ $p(3)=(3)^{3}+2(3)^{2}+3 k+3$ $=27+18+3 \mathrm{k}+3$ $=48+3 k$ Since, that the reminder is 21 $\therefore 3 \mathrm{k}+48=21$ $\Rightarrow 3 \mathrm{k}=-27$ $\Rightarrow...
If two zeroes of the polynomial are and , find its other two zeroes.
$p(x)=2 x^{4}-3 x^{3}-3 x^{2}+6 x-2$ and the two zeroes, $\sqrt{2}$ and $-\sqrt{2}$ the polynomial is $(x+\sqrt{2})(x-\sqrt{2})=x^{2}-2$. dividing $p(x)$ by $\left(x^{2}-2\right)$ $2 x^{4}-3 x^{3}-3...
If one zero of the polynomial is 3 , find the other two zeroes.
$p(x)=x^{3}-6 x^{2}+11 x-6$ and its factor, $x+3$ dividing $p(x)$ by $(x-3)$ $x^{3}-6 x^{2}+11 x-6=(x-3)\left(x^{2}-3 x+2\right)$ $$ \begin{aligned} &=(x-3)\left[\left(x^{2}-(2+1)...
Show that the polynomial has no zero.
$\operatorname{Let} t=\mathrm{x}^{2}$ $f(t)=t^{2}+4 t+6$ to find the zeroes, we will equate $\mathrm{f}(\mathrm{t})=0$ $\Rightarrow t^{2}+4 t+6=0$ $$ \text { Now, t } \begin{aligned} &=\frac{-4 \pm...
If are the zeroes of the polynomial such that , find the value of .
$x^{2}-5 x+k$ The co-efficients are $\mathrm{a}=1, \mathrm{~b}=-5$ and $\mathrm{c}=\mathrm{k}$. $\therefore \boldsymbol{\alpha}+\boldsymbol{\beta}=\frac{-\boldsymbol{b}}{\boldsymbol{a}}$...
If are the zeroes of the polynomial , find the value of
Given: p(x)=6x3+3x2-5x+1 \text { Given: } p(x)=6 x^{3}+3 x^{2}-5 x+1 =6x3-(-3)x2+(-5)x-1 =6 \mathrm{x}^{3}-(-3) \mathrm{x}^{2}+(-5) \mathrm{x}-1 Comparing the polynomial with...
Show that is a factor of .
$f(x)=x^{3}+4 x^{2}+x-6$ $f(-2)=(-2)^{3}+4(-2)^{2}+(-2)-6$ $=-8+16-2-6$ $=0$ $\therefore(x+2)$ is a factor of $f(x)=x^{3}+4 x^{2}+x-6$.
If the zeroes of the polynomial are , a and , find the values of a and .
$=x^{3}-3 x^{2}+x+1$ and its roots are $(a-b)$, a and $(a+b)$. Comparing the given polynomial with $\mathrm{Ax}^{3}+\mathrm{Bx}^{2}+\mathrm{Cx}+\mathrm{D}$, we have: $A=1, B=-3, C=1$ and $D=1$...
Find a quadratic polynomial whose zeroes are 2 and .
Since, the two roots of the polynomial are 2 and $-5$. Let $\boldsymbol{\alpha}=2$ and $\boldsymbol{\beta}=-5$ the sum of the zeroes, $\boldsymbol{\alpha}+\boldsymbol{\beta}=2+(-5)=-3$ Product of...
If one zero of the polynomial is the reciprocal of the other, find the value of a.
$(a+9) x^{2}-13 x+6 a=0$ Here, $A=\left(a^{2}+9\right), B=13$ and $C=6$ a Let $\boldsymbol{\alpha}$ and $\frac{\mathbf{1}}{\boldsymbol{\alpha}}$ be the two zeroes. Then, product of the zeroes...
Find the zeroes of the polynomial .
$p(x)=x^{2}+2 x-195$ Let $p(x)=0$ $\Rightarrow x^{2}+(15-13) x-195=0$ $\Rightarrow x^{2}+15 x-13 x-195=0$ $\Rightarrow \mathrm{x}(\mathrm{x}+15)-13(\mathrm{x}+15)=0$ $\Rightarrow(x+15)(x-13)=0$...
It is given that the difference between the zeroes of is 4 and . Then, (a) (b) (c) (d)
The correct option is option (c) $\frac{5}{2}$ Let the zeroes of the polynomial be $\boldsymbol{\alpha}$ and $\boldsymbol{\alpha}+4$ $p(x)=4 x^{2}-8 k x+9$ Comparing the given polynomial with $a...
The zeroes of the polynomial are (a) (b) (c) (d) 3,1
(c) $3,-1$ Here, $p(x)=x^{2}-2 x-3$ Let $x^{2}-2 x-3=0$ $\Rightarrow x^{2}-(3-1) x-3=0$ $\Rightarrow x^{2}-3 x+x-3=0$ $\Rightarrow \mathrm{x}(\mathrm{x}-3)+1(\mathrm{x}-3)=0$...
On dividing a polynomial by a non-zero polynomial be the quotient and be the remainder, then , where (a) always (b) always (c) either or (d)
The correct option is (c) either $\mathrm{r}(\mathrm{x})=0$ or $\operatorname{deg} \mathrm{r}(\mathrm{x})<\operatorname{deg} \mathrm{g}(\mathrm{x})$ By division algorithm on polynomials, either...
If be the zeroes of the polynomial such that , then ? (a) 3 (b) (c) -2 (d) 2
The correct option is option (d) 2 $\alpha$ and $\beta$ are the zeroes of $2 \mathrm{x}^{2}+5 \mathrm{x}+\mathrm{k}$, we have: $\alpha+\beta=\frac{-5}{2}$ and $\alpha \beta=\frac{k}{2}$ it is given...
If one of the zeroes of the cubic polynomial is , then the product of the other two zeroes is (a) (b) (c) (d)
The correct option is option (c) $1-\mathrm{a}+\mathrm{b}$ $-1$ is a zero of $x^{3}+a x^{2}+b x+c$, we have $(-1)^{3}+a \times(-1)^{2}+b \times(-1)+c=0$ $\Rightarrow a-b+c+1=0$ $\Rightarrow...
If two of the zeroes of the cubic polynomial are 0 , then the third zero is (a) (b) (c) (d)
The correct option is option (a) $\frac{-b}{a}$ $\alpha, 0$ and 0 be the zeroes of $a x^{3}+b x^{2}+c x+d=0$ Then the sum of zeroes $=\frac{-b}{a}$ $\Rightarrow \alpha+0+0=\frac{-b}{a}$ $\Rightarrow...
If be the zeroes of the polynomial such that and , then (a) (b) (c) (d) none of these
The correct option is option (c) $x^{3}-3 x^{2}-10 x+24$ $\alpha, \beta$ and $\gamma$ are the zeroes of polynomial $p(x)$. $(\alpha+\beta+\gamma)=3,(\alpha \beta+\beta \gamma+\gamma \alpha)=-10$ and...
If be the zeroes of the polynomial , then (a) (b) 1 (c) (d) 30
The correct option is option (a) $-1$ It is given that $\alpha, \beta$ and $\gamma$ are the zeroes of $x^{3}-6 x^{2}-x+30$ $\therefore(\alpha \beta+\beta \gamma+\gamma \alpha)=\frac{\text {...
If are the zeroes of the polynomial , then ? (a) 3 (b) (c) 12 (d)
The correct option is option (b) $-3$ $\alpha$ and $\beta$ be the zeroes of $x^{2}+6 x+2$, we have: $\alpha+\beta=-6$ and $\alpha \beta=2$...
If and 3 are the zeroes of the quadratic polynomial , then (a) (b) (c) (d)
The correct option is option (c) $a=-2, b=-6$ $-2$ and 3 are the zeroes of $x^{2}+(a+1) x+b$. $(-2)^{2}+(a+1) \times(-2)+b=0 \Rightarrow 4-2 a-2+b=0$ $\Rightarrow b-2 a=-2$ ….(1) $3^{2}+(a+1) \times...
If one zero of the quadratic polynomial is 2 , then the value of is (a) (b) (c) (d)
The correct option is option (d) $\frac{-6}{5}$ Since 2 is a zero of $k x^{2}+3 x+k$, we have: $\mathrm{k} \times(2)^{2}+3(2)+\mathrm{k}=0$ $\Rightarrow 4 \mathrm{k}+\mathrm{k}+6=0$ $\Rightarrow 5...
If and are the zeros of , then the value of is (a) 5 (b) (c) 8 (d)
The correct option is option (b) $-5$ $\alpha$ and $\beta$ be the zeroes of $\mathrm{x}^{2}+5 \mathrm{x}+8$. If $\alpha+\beta$ is the sum of the roots and $\alpha \beta$ is the product, then the...
The zeroes of the quadratic polynomial are (a) both positive (b) both negative (c) one positive and one negative (d) both equal
The correct option is option (b) both negative $\alpha$ and $\beta$ be the zeroes of $\mathrm{x}^{2}+88 \mathrm{x}+125$ Then $\alpha+\beta=-88$ and $\alpha \times \beta=125$ This can only happen...
A quadratic polynomial whose zeroes are and , is (a) (b) (c) (d)
The correct option is option (d) $x^{2}-\frac{1}{10} x-\frac{3}{10}$ Since, the zeroes are $\frac{3}{5}$ and $\frac{-1}{2}$ $\alpha=\frac{3}{5}$ and $\beta=\frac{-1}{2}$ sum of the zeroes,...
A quadratic polynomial whose zeroes are 5 and -3, is (a) (b) (c) (d) none of these
(c) $x^{2}-2 x-15$ Since, the zeroes are 5 and $-3$. $\alpha=5$ and $\beta=-3$ Therefore, sum of the zeroes, $\alpha+\beta=5+(-3)=2$ product of the zeroes, $\alpha \beta=5 \times(-3)=-15$ The...
The sum and product of the zeroes of a quadratic polynomial are 3 and respectively. The quadratic polynomial is (a) (b) (c) (d)
The correct option is option (c) $x^{2}-3 x-10$ Sum of zeroes, $\alpha+\beta=3$ Also, product of zeroes, $\alpha \beta=-10$ $\therefore$ Required polynomial $=\mathrm{x}^{2}-(\alpha+\beta)+\alpha...
The zeros of the polynomial are (a) (a) (c) (d) none of these
The correct option is option (a) $\frac{2}{3}, \frac{-1}{7}$ $f(x)=7 x^{2}-\frac{11}{3} x-\frac{2}{3}=0$ $\Rightarrow 21 \mathrm{x}^{2}-11 \mathrm{x}-2=0$ $\Rightarrow 21 \mathrm{x}^{2}-14...
The zeros of the polynomial are (a) (b) (c) (d) none of these
The correct option is option (b) $\frac{-3}{2}, \frac{4}{3}$ $f(x)=x^{2}+\frac{1}{6} x-2=0$ $\Rightarrow 6 \mathrm{x}^{2}+\mathrm{x}-12=0$ $\Rightarrow 6 x^{2}+9 x-8 x-12=0$ $\Rightarrow 3 x(2...
The zeroes of the polynomial are (a) (b) (c) (d) none of these
The correct option is option (c) $\frac{3}{\sqrt{2}}, \frac{\sqrt{2}}{4}$ $f(x)=4 x^{2}+5 \sqrt{2} x-3=0$ $\Rightarrow 4 x^{2}+6 \sqrt{2} x-\sqrt{2} x-3=0$ $\Rightarrow 2 \sqrt{2}...
The zeroes of the polynomial are (a) (b) (c) (d)
The correct option is option (b) $3 \sqrt{2},-2 \sqrt{2}$ $f(x)=x^{2}-\sqrt{2} x-12=0$ $\Rightarrow x^{2}-3 \sqrt{2} x+2 \sqrt{2} x-12=0$ $\Rightarrow x(x-3 \sqrt{2})+2 \sqrt{2}(x-3 \sqrt{2})=0$...
The Zeroes of the polynomial are (a) (b) (c) (d) 3,1
The correct option is option (c) $3,-1$ Let $f(x)=x^{2}-2 x-3=0$ $=x^{2}-3 x+x-3=0$ $=x(x-3)+1(x-3)=0$ $=(x-3)(x+1)=0$ $\Rightarrow \mathrm{x}=3$ or $\mathrm{x}=-1$
Which of the following is not a polynomial? (a) (b) (c) (d)
The correct option is option (d) $x+\frac{3}{x}$ is not a polynomial. It is because in the second term, the degree of $x$ is $-1$ and an expression with a negative degree is not a polynomial.
Which of the following is a polynomial? (a) (b) (c) (d) None of these
The correct option is option (d) none of these A polynomial in $x$ of degree $n$ is an expression of the form $p(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \ldots+$ $a_{n} x^{n}$, where $a_{n} \neq 0$.
If the zeroes of the polynomial are , a and , find the values of a and b.
using the relationship between the zeroes of he quadratic polynomial. Sum of zeroes $=\frac{-\left(\text { coef ficient of } x^{2}\right)}{\text { coefficient of } x^{3}}$ $\therefore...
If are the zeroes of the polynomial , then .
using the relationship between the zeroes of the quadratic polynomial. We have Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes...
If are the zeroes of the polynomial , then
using the relationship between the zeroes of he quadratic polynomial. Sum of zeroes $=\frac{-\text { (coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
If and are the zeros of the polynomial find the value of
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
If are the zeroes of the polynomial , then ? (a) 3 (b) (c) 12 (d)
The correct option is option (b) $-3$ Since $\alpha$ and $\beta$ be the zeroes of $x^{2}+6 x+2$, we have: $\alpha+\beta=-6$ and $\alpha \beta=2$...
If one zero of be the reciprocal of the other, then ? (a) 3 (b) (c) (d)
The correct option is option (a) $\mathrm{k}=3$ Let $\alpha$ and $\frac{1}{\alpha}$ be the zeroes of $3 x^{2}-8 x+k$. Then the product of zeroes $=\frac{k}{3}$ $\Rightarrow \alpha \times...
If one zero of the quadratic polynomial is , then the value of is (a) (b) (c) (d)
The correct option is option (b) $\frac{5}{4}$ Since $-4$ is a zero of $(k-1) x^{2}+k x+1$, we have: $(k-1) \times(-4)^{2}+k \times(-4)+1=0$ $\Rightarrow 16 \mathrm{k}-16-4 \mathrm{k}+1=0$...
If one zero of the quadratic polynomial is 2 , then the value of is (a) (b) (c) (d)
The correct option is option (d) $\frac{-6}{5}$ Since 2 is a zero of $k x^{2}+3 x+k$, we have: $\mathrm{k} \times(2)^{2}+3(2)+\mathrm{k}=0$ $\Rightarrow 4 \mathrm{k}+\mathrm{k}+6=0$ $\Rightarrow 5...
If and are the zeroes of , then the value of is (a) (b) (c) (d)
The correct option is option (c) $\frac{-9}{2}$ Since, $\alpha$ and $\beta$ be the zeroes of $2 \mathrm{x}^{2}+5 \mathrm{x}-9$. If $\alpha+\beta$ are the zeroes, then $\mathrm{x}^{2}-(\alpha+\beta)...
The zeroes of the quadratic polynomial are (a) both positive (b) both negative (c) one positive and one negative (d) both equal
The correct option is option (b) both negative Let $\alpha$ and $\beta$ be the zeroes of $\mathrm{x}^{2}+88 \mathrm{x}+125$ Then $\alpha+\beta=-88$ and $\alpha \times \beta=125$ This can only happen...
A quadratic polynomial whose zeroes are 5 and -3, is (a) (b) (c) (d) none of these
The correct option is option (c) $x^{2}-2 x-15$ Since, the zeroes are 5 and $-3$. Let $\alpha=5$ and $\beta=-3$ sum of the zeroes, $\alpha+\beta=5+(-3)=2$ Also, product of the zeroes, $\alpha...
The zeros of the polynomial are (a) (a) (c) (d) none of these
The correct option is option (a) $\frac{2}{3}, \frac{-1}{7}$ Let $f(x)=7 x^{2}-\frac{11}{3} x-\frac{2}{3}=0$ $\Rightarrow 21 \mathrm{x}^{2}-11 \mathrm{x}-2=0$ $\Rightarrow 21 \mathrm{x}^{2}-14...
The zeroes of the polynomial are (a) (b) (c) (d) none of these
The correct option is option (c) $\frac{3}{\sqrt{2}}, \frac{\sqrt{2}}{4}$ Let $f(x)=4 x^{2}+5 \sqrt{2} x-3=0$ $\Rightarrow 4 x^{2}+6 \sqrt{2} x-\sqrt{2} x-3=0$ $\Rightarrow 2 \sqrt{2}...
The Zeroes of the polynomial are (a) (b) (c) (d) 3,1
The correct option is option(c) $3,-1$ Let $f(x)=x^{2}-2 x-3=0$ $=x^{2}-3 x+x-3=0$ $=x(x-3)+1(x-3)=0$ $=(x-3)(x+1)=0$ $\Rightarrow \mathrm{x}=3$ or $\mathrm{x}=-1$
Which of the following is not a polynomial? (a) (b) (c) (d)
The correct option is option(d) $x+\frac{3}{x}$ is not a polynomial. It is because in the second term, the degree of $x$ is $-1$ and an expression with a negative degree is not a polynomial.
Which of the following is a polynomial? (a) (b) (c) (d) None of these
The correct option is option(d) none of these A polynomial in $x$ of degree $n$ is an expression of the form $p(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \ldots+$ $a_{n} x^{n}$, where $a_{n} \neq 0$.
If the zeroes of the polynomial are , a and , find the values of a and b.
using the relationship between the zeroes of he quadratic polynomial. Sum of zeroes $=\frac{-\left(\text { coef ficient of } x^{2}\right)}{\text { coefficient of } x^{3}}$ $\therefore...
If are the zeroes of the polynomial , then
using the relationship between the zeroes of he quadratic polynomial. => Sum of zeroes $=\frac{-\text { (coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text...
If and are the zeros of the polynomial find the value of
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
If are the zeroes of the polynomial such that , find the value of
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
Find the zeroes of the quadratic polynomial .
For finding the zeroes of the quadratic polynomial we will equate $f(x)$ to 0 Hence, the zeroes of the quadratic polynomial $f(x)=4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}$ are $-\frac{2}{\sqrt{3}}$ or...
Find the zeroes of the quadratic polynomial .
To find the zeroes of the quadratic polynomial we will equate $\mathrm{f}(\mathrm{x})$ to 0 Hence, the zeroes of the quadratic polynomial $f(x)=6 x^{2}-3$ are...
Find the sum of the zeros and the product of zeros of a quadratic polynomial, are and respectively. Write the polynomial.
We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula $\mathrm{x}^{2}-($ sum of the zeroes $) \mathrm{x}+$ product of zeroes $\Rightarrow...
State Division Algorithm for Polynomials.
"If $\mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ are two polynomials such that degree of $\mathrm{f}(\mathrm{x})$ is greater than degree of $\mathrm{g}(\mathrm{x})$ where...
If and be the zeroes of the polynomial write the value of .
using the relationship between the zeroes of he quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
If is divisible by , write the value of a and .
Equating $\mathrm{x}^{2}-\mathrm{x}$ to 0 to find the zeroes, we will get x(x-1)=0 x(x-1)=0 ⇒x=0 or x-1=0 \Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}-1=0...
If and are zeros of the polynomial write the value of a.
using the relationship between the zeroes of the quadratic polynomial.$$ \begin{aligned} &\text { Sum of zeroes }=\frac{-\left(\text { coefficient of } x^{2}\right)}{\text { coefficient of } x^{3}}...
If the product of the zero of the polynomial is 3 . Find the value of .
using the relationship between the zeroes of he quadratic polynomial. Product of zeroes $=\frac{\text { constant term }}{\text { coefficient of } x^{2}}$ $\Rightarrow 3=\frac{k}{1}$ $\Rightarrow...
If the sum of the zeros of the quadratic polynomial is 1 write the value of .
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ $\Rightarrow 1=\frac{-(-3)}{k}$...
Write the zeros of the polynomial .
$f(x)=x^{2}-x-6$ $=x^{2}-3 x+2 x-6$ $=x(x-3)+2(x-3)$ $=(x-3)(x+2)$ $f(x)=0 \Rightarrow(x-3)(x+2)=0$ $$ \begin{aligned} &\Rightarrow(x-3)=0 \text { or }(x+2)=0 \\ &\Rightarrow x=3 \text { or } x=-2...
If is a zero of the polynomial then find the value of .
$x=-2$ is one zero of the polynomial $3 x^{2}+4 x+2 k$ Therefore, it will satisfy the above polynomial. Now, we have $3(-2)^{2}+4(-2) 1+2 k=0$ $\Rightarrow 12-8+2 k=0$ $\Rightarrow...
If one zero of the quadratic polynomial is 2 , then find the value of .
$x=2$ is one zero of the quadratic polynomial $k x^{2}+3 x+k$ Therefore, it will satisfy the above polynomial. $k(2)^{2}+3(2)+k=0$ $\Rightarrow 4 \mathrm{k}+6+\mathrm{k}=0$ $\Rightarrow 5...
Find are the zeros of polynomial and then write the polynomial.
If the zeroes of the quadratic polynomial are $\alpha$ and $\beta$ then the quadratic polynomial can be found as $\mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$ $\ldots \ldots(1)$...
Find the zeroes of the polynomial
$f(x)=x^{2}-3 x-m(m+3)$ adding and subtracting $\mathrm{mx}$, $f(x)=x^{2}-m x-3 x+m x-m(m+3)$ $=x[x-(m+3)]+m[x-(m+3)]$ $=[x-(m+3)](x+m)$ $f(x)=0 \Rightarrow[x-(m+3)](x+m)=0$...
Find all the zeroes of polynomial , it being given that two of its zeroes are and .
$f(x)=2 x^{4}-11 x^{3}+7 x^{2}+13 x-7$. Since $(3+\sqrt{2})$ and $(3-\sqrt{2})$ are the zeroes of $f(x)$ it follows that each one of $(x+3+\sqrt{2})$ and $(x+3-\sqrt{2})$ is a factor of $f(x)$...
Find all the zeroes of , it is being given that two of its zeroes are and .
The given polynomial is $f(x)=2 x^{4}-3 x^{3}-5 x^{2}+9 x-3$ Since $\sqrt{3}$ and $-\sqrt{3}$ are the zeroes of $f(x)$, it follows that each one of $(x-\sqrt{3})$ and $(x+\sqrt{3})$ is a factor of...
If 2 and are two zeroes of the polynomial , find all the zeroes of the given polynomial.
Let $f(x)=x^{4}+x^{3}-34 x^{2}-4 x+120$ Since 2 and $-2$ are the zeroes of $f(x)$, it follows that each one of $(x-2)$ and $(x+2)$ is a factor of $f(x)$ Consequently,...
If 3 and are two zeroes of the polynomial , find all the zeroes of the given polynomial.
Since 3 and $-3$ are the zeroes of $f(x)$, it follows that each one of $(x+3)$ and $(x-3)$ is a factor of $f(x)$ Consequently, $(x-3)(x+3)=\left(x^{2}-9\right)$ is a factor of $f(x)$. On dividing...
If 1 and are two zeroes of the polynomial , find its third zero.
Let $f(x)=x^{3}-4 x^{2}-7 x+10$ Since 1 and $-2$ are the zeroes of $f(x)$, it follows that each one of $(x-1)$ and $(x+2)$ is a factor of $f(x)$ Consequently, $(x-1)(x+2)=\left(x^{2}+x-2\right)$ is...
It is given that is one of the zeroes of the polynomial . Find all the zeroes of the given polynomial.
Let $f(x)=x^{3}+2 x^{2}-11 x-12$ Since $-1$ is a zero of $f(x),(x+1)$ is a factor of $f(x)$. On dividing $\mathrm{f}(\mathrm{x})$ by $(\mathrm{x}+1)$, we get $$ \begin{aligned} &f(x)=x^{3}+2...
On dividing is divided by a polynomial , the quotient and remainder are and respectively. Find
using division rule, Dividend $=$ Quotient $\times$ Divisor $+$ Remainder $\therefore 3 x^{3}+x^{2}+2 x+5=(3 x-5) g(x)+9 x+10$ $\Rightarrow 3 x^{3}+x^{2}+2 x+5-9 x-10=(3 x-5) g(x)$ $\Rightarrow 3...
If is divided by
Quotient $q(x)=x-\overline{3}$ Remainder $r(x)=7 x-9$ 7. If $f(x)=x^{4}-3 x^{2}+4 x+5$ is divided by $g(x)=x^{2}-x+1$
Find a cubic polynomial whose zeroes are and .
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as $x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c$ Let $a=\frac{1}{2}, b=1$ and $c=-3$ Substituting the values...
Find the quadratic polynomial, sum of whose zeroes is and their product is .
Quadratic equation can be found if we know the sum of the roots and product of the roots by using the formula: $\mathrm{x}^{2}-($ Sum of the roots) $\mathrm{x}+$ Product of roots $=0$ $\Rightarrow...
Find the quadratic polynomial, sum of whose zeroes is 0 and their product is . Hence, find the zeroes of the polynomial.
Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $f(x)$. Then $(\alpha+\beta)=0$ and $\alpha \beta=-1$ $\therefore f(x)=x^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$ $\Rightarrow...
Find the zeroes of the quadratic polynomial and verify the relation between the zeroes and the coefficients.
$$ \begin{aligned} \mathrm{f}(\mathrm{x}) &=2 \mathrm{x}^{2}-11 \mathrm{x}+15 \\ &=2 \mathrm{x}^{2}-(6 \mathrm{x}+5 \mathrm{x})+15 \\ &=2 \mathrm{x}^{2}-6 \mathrm{x}-5 \mathrm{x}+15 \\ =& 2...