$p(x)=x^{3}+2 x^{2}+k x+3$ $p(3)=(3)^{3}+2(3)^{2}+3 k+3$ $=27+18+3 \mathrm{k}+3$ $=48+3 k$ Since, that the reminder is 21 $\therefore 3 \mathrm{k}+48=21$ $\Rightarrow 3 \mathrm{k}=-27$ $\Rightarrow...
$p(x)=2 x^{4}-3 x^{3}-3 x^{2}+6 x-2$ 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...
$p(x)=x^{3}-6 x^{2}+11 x-6$ and its factor, $x+3$ dividing $p(x)$ by $(x-3)$ $x^{3}-6 x^{2}+11 x-6=(x-3)\left(x^{2}-3 x+2\right)$ $$ \begin{aligned} &=(x-3)\left[\left(x^{2}-(2+1)...
$\operatorname{Let} t=\mathrm{x}^{2}$ $f(t)=t^{2}+4 t+6$ to find the zeroes, we will equate $\mathrm{f}(\mathrm{t})=0$ $\Rightarrow t^{2}+4 t+6=0$ $$ \text { Now, t } \begin{aligned} &=\frac{-4 \pm...
$x^{2}-5 x+k$ The co-efficients are $\mathrm{a}=1, \mathrm{~b}=-5$ and $\mathrm{c}=\mathrm{k}$. $\therefore \boldsymbol{\alpha}+\boldsymbol{\beta}=\frac{-\boldsymbol{b}}{\boldsymbol{a}}$...
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...
$f(x)=x^{3}+4 x^{2}+x-6$ $f(-2)=(-2)^{3}+4(-2)^{2}+(-2)-6$ $=-8+16-2-6$ $=0$ $\therefore(x+2)$ is a factor of $f(x)=x^{3}+4 x^{2}+x-6$.
$=x^{3}-3 x^{2}+x+1$ 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 $D=1$...
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...
$(a+9) x^{2}-13 x+6 a=0$ 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...
$p(x)=x^{2}+2 x-195$ Let $p(x)=0$ $\Rightarrow x^{2}+(15-13) x-195=0$ $\Rightarrow x^{2}+15 x-13 x-195=0$ $\Rightarrow \mathrm{x}(\mathrm{x}+15)-13(\mathrm{x}+15)=0$ $\Rightarrow(x+15)(x-13)=0$...
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...
(c) $3,-1$ Here, $p(x)=x^{2}-2 x-3$ Let $x^{2}-2 x-3=0$ $\Rightarrow x^{2}-(3-1) x-3=0$ $\Rightarrow x^{2}-3 x+x-3=0$ $\Rightarrow \mathrm{x}(\mathrm{x}-3)+1(\mathrm{x}-3)=0$...
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...
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...
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...
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...
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...
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 {...
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 $\alpha \beta=2$...
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...
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...
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 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...
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,...
(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 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 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 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 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 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 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 $\mathrm{x}=-1$
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.
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 $a_{n} \neq 0$.
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...
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...
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 {...
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 {...
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 $\alpha \beta=2$...
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...
The correct option is option (b) $\frac{5}{4}$ Since $-4$ is a zero of $(k-1) x^{2}+k x+1$, we have: $(k-1) \times(-4)^{2}+k \times(-4)+1=0$ $\Rightarrow 16 \mathrm{k}-16-4 \mathrm{k}+1=0$...
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...
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 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...
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 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 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 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 $\mathrm{x}=-1$
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.
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 $a_{n} \neq 0$.
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...
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...
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 {...
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 {...
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...
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...
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...
"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...
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 {...
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...
using the relationship between the zeroes of the quadratic polynomial.$$ \begin{aligned} &\text { Sum of zeroes }=\frac{-\left(\text { coefficient of } x^{2}\right)}{\text { coefficient of } x^{3}}...
using the relationship between the zeroes of he quadratic polynomial. Product of zeroes $=\frac{\text { constant term }}{\text { coefficient of } x^{2}}$ $\Rightarrow 3=\frac{k}{1}$ $\Rightarrow...
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ $\Rightarrow 1=\frac{-(-3)}{k}$...
$f(x)=x^{2}-x-6$ $=x^{2}-3 x+2 x-6$ $=x(x-3)+2(x-3)$ $=(x-3)(x+2)$ $f(x)=0 \Rightarrow(x-3)(x+2)=0$ $$ \begin{aligned} &\Rightarrow(x-3)=0 \text { or }(x+2)=0 \\ &\Rightarrow x=3 \text { or } x=-2...
$x=-2$ 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...
$x=2$ 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...
If the zeroes of the quadratic polynomial are $\alpha$ and $\beta$ then the quadratic polynomial can be found as $\mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$ $\ldots \ldots(1)$...
$f(x)=x^{2}-3 x-m(m+3)$ adding and subtracting $\mathrm{mx}$, $f(x)=x^{2}-m x-3 x+m x-m(m+3)$ $=x[x-(m+3)]+m[x-(m+3)]$ $=[x-(m+3)](x+m)$ $f(x)=0 \Rightarrow[x-(m+3)](x+m)=0$...
$f(x)=2 x^{4}-11 x^{3}+7 x^{2}+13 x-7$. 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 $f(x)$...
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...
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,...
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...
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...
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 $$ \begin{aligned} &f(x)=x^{3}+2...
using division rule, Dividend $=$ Quotient $\times$ Divisor $+$ Remainder $\therefore 3 x^{3}+x^{2}+2 x+5=(3 x-5) g(x)+9 x+10$ $\Rightarrow 3 x^{3}+x^{2}+2 x+5-9 x-10=(3 x-5) g(x)$ $\Rightarrow 3...
Quotient $q(x)=x-\overline{3}$ Remainder $r(x)=7 x-9$ 7. If $f(x)=x^{4}-3 x^{2}+4 x+5$ is divided by $g(x)=x^{2}-x+1$
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as $x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c$ Let $a=\frac{1}{2}, b=1$ and $c=-3$ Substituting the values...
Quadratic equation can be found if we know the sum of the roots and product of the roots by using the formula: $\mathrm{x}^{2}-($ Sum of the roots) $\mathrm{x}+$ Product of roots $=0$ $\Rightarrow...
Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $f(x)$. Then $(\alpha+\beta)=0$ and $\alpha \beta=-1$ $\therefore f(x)=x^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$ $\Rightarrow...
$$ \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...