is divided by 
.$p(x)=3 x^{4}+5 x^{3}-7 x^{2}+2 x+2$ Dividing $p(x)$ by $\left(x^{2}+3 x+1\right)$, we have $$ \begin{gathered} \left.x^{2}+3 x+1\right) \begin{array}{l} 3 x^{4}+5 x^{3}-7 x^{2}+2 x+2 \\ 3 x^{4}+9...
is divided by 
$p(x)=x^{3}+3 x^{2}-5 x+4$ $p(2)=2^{3}+3\left(2^{2}\right)-5(2)+4$ $=8+12-10+4$ $=14$
$\boldsymbol{\alpha}, \boldsymbol{\beta}$ and $\gamma$ are the zeroes of the required polynomial. $\boldsymbol{\alpha}+\boldsymbol{\beta}+\boldsymbol{\gamma}=3+5+(-2)=6$ $\boldsymbol{\alpha}...
are the zeros of 
is equal 
then 
? (a) 
(b) 
(c) 
(d) 
The correct option is option (c) $\frac{2}{3}$ $p(x)=x^{2}-2 x+3 k$ Comparing the given polynomial with $a x^{2}+b x+c$, we get: $a=1, b=-2$ and $c=3 k$ It is given that $\boldsymbol{\alpha}$ and...
be the zeroes of the polynomial 
, then the values of 
(a) 
(b) 1 (c) 
(d) 3The correct option is option (a) $-1$ $p(x)=x^{3}-6 x^{2}-x+3$ Comparing the given polynomial with $\mathrm{x}^{3}-(\boldsymbol{\alpha}+\boldsymbol{\beta}+\gamma) \mathrm{x}^{2}+(\boldsymbol{\alpha}...
is a linear polynomial. (b) 
is a binomial (c) 
is a monomial (d) 
is a monomialThe correct option is option (d) $5 \mathrm{x}^{2}$ is a monomial. $5 \mathrm{x}^{2}$ consists of one term only. So, it is a monomial.
is 0 , then the product of the other two zeroes is (a) 
(b) 
(c) 0 (d) 
The correct option is option (b) $\frac{c}{a}$ $\alpha, \beta$ and 0 be the zeroes of $a x^{3}+b x^{2}+c x+d$. Then, sum of the products of zeroes taking two at a time is given by $(\alpha...
be the zeroes of the polynomial 
, then 
? (a) 
(b) 3 (c) 
(d) 
The correct option is option (a) $-3$ $\alpha, \beta$ and $\gamma$ are the zeroes of $2 \mathrm{x}^{3}+\mathrm{x}^{2}-13 \mathrm{x}+6$, we have: $\alpha \beta \gamma=\frac{-(\text { constant term...
be the reciprocal of the other, then 
? (a) 3 (b) (c) (d) The correct option is option (a) $\mathrm{k}=3$ $\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...
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$ $(k-1) \times(-4)^{2}+k \times(-4)+1=0$ $\Rightarrow 16 \mathrm{k}-16-4 \mathrm{k}+1=0$ $\Rightarrow 12...
and 
are the zeroes of 
, then the value of 
is (a) 
(b) 
(c) 
(d) 
The correct option is option (c) $\frac{-9}{2}$ $\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)...
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 {...
is equal to the product of its zeroes, then ? (a) (b) (c) (d) The correct option is option (d) $\frac{-2}{3}$ Assuming $\alpha$ and $\beta$ be the zeroes of $\mathrm{kx}^{2}+2 \mathrm{x}+3 \mathrm{k}$. Then $\alpha+\beta=\frac{-2}{k}$ and $\alpha \beta=3$...
and 3 are the zeroes of the quadratic polynomial 
, then (a) 
(b) 
(c) 
(d) 
The correct option is option (c) $a=-2, b=-6$ Since, $-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) Also,...
and are the zeros of 
, then the value of 
is (a) 5 (b) (c) 8 (d) 
The correct option is option (b) $-5$ Since, $\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...
and , is (a) 
(b) 
(c) 
(d) 
The correct option is option (d) $x^{2}-\frac{1}{10} x-\frac{3}{10}$ the zeroes are $\frac{3}{5}$ and $\frac{-1}{2}$ Let $\alpha=\frac{3}{5}$ and $\beta=\frac{-1}{2}$ sum of the zeroes,...
respectively. The quadratic polynomial is (a) 
(b) 
(c) 
(d) 
The correct option is option (c) $x^{2}-3 x-10$ Since, sum of zeroes, $\alpha+\beta=3$ Also, product of zeroes, $\alpha \beta=-10$ $\therefore$ Required polynomial...
are (a) 
(b) 
(c) 
(d) none of theseThe correct option is option (b) $\frac{-3}{2}, \frac{4}{3}$ Let $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...
are (a) 
(b) 
(c) 
(d) 
The correct option is option (b) $3 \sqrt{2},-2 \sqrt{2}$ Let $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$...
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...
is a factor of 
, then find the value of 
$(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ $x+a=0$ $\Rightarrow \mathrm{X}=-\mathrm{a}$ Since, $(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ Hence, It will satisfy the above polynomial...
is 1 , then find the value of a.$x=1$ is one zero of the polynomial $a x^{2}-3(a-1) x-1$ Therefore, it will satisfy the above polynomial. Now, we have $a(1)^{2}-(a-1) 1-1=0$ $\Rightarrow a-3 a+3-1=0$ $\Rightarrow-2 \mathrm{a}=-2$...
is a zero of the polynomial 
is , then find the value of 
.$x=-4$ is one zero of the polynomial $x^{2}-x-(2 k+2)$ Therefore, it will satisfy the above polynomial. Now, we have $(-4)^{2}-(-4)-(2 k+2)=0$ $\Rightarrow 16+4-2 \mathrm{k}-2=0$ $\Rightarrow 2...
$f(x)=x^{2}+x-p(p+1)$ adding and subtracting $\mathrm{px}$, we get $f(x)=x^{2}+p x+x-p x-p(p+1)$ $=x^{2}+(p+1) x-p x-p(p+1)$ $=x[x+(p+1)]-p[x+(p+1)]$ $=[x+(p+1)](x-p)$ $f(x)=0$...
Is 
, write the other zero.Let the other zeroes of $x^{2}-4 x+1$ be a (using the relationship between the zeroes of the quadratic polynomial) sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coef ficient of }...
if two of its zeroes are 
and 
.The given polynomial is $f(x)=x^{4}+4 x^{3}-2 x^{2}-20 x-15$. Since $(x-\sqrt{5})$ and $(x+\sqrt{5})$ are the zeroes of $f(x)$ it follows that each one of $(x-\sqrt{5})$ and $(x$ $+\sqrt{5})$ is a...
by 
$-6 x^{3}+x^{2}+20 x+8$ and $g(x)$ as $-3 x^{2}+5 x+2$ Quotient $=2 \mathrm{x}+3$ Remainder $=x+2$ By using division rule, we have Dividend $=$ Quotient $\times$ Divisor $+$ Remainder $\therefore-6...
is a factor of 
.Let $f(x)=2 x^{4}+3 x^{3}-2 x^{2}-9 x-12$ and $g(x)$ as $x^{2}-3$
is divided by 
.$f(x)$ as $x^{4}+0 x^{3}+0 x^{2}-5 x+6$ and $g(x) a s-x^{2}+2$ Quotient $q(x)=-x^{2}-2$ Remainder $\mathrm{r}(\mathrm{x})=-5 \mathrm{x}+10$
and 
respectively.sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as $x^{3}-($ sum of the zeroes $) x^{2}+($ sum of the...
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=2, b=-3$ and $c=4$ Substituting the values in 1 , we...
and are the zeroes of the cubic polynomial 
and verify the relation between its zeroes and coefficients.p(x)=3x3-10x2-27x+10 p(x)=\left(3 x^{3}-10 x^{2}-27 x+10\right) p(5)=3×53-10×52-27×5+10=(375-250-135+10)=0 p(5)=\left(3 \times 5^{3}-10 \times 5^{2}-27 \times...
are the zeros of the cubic polynomial 
and verify the relation between it zeros and coefficients.The given polynomial is $p(x)=\left(x^{3}-2 x^{2}-5 x+6\right)$ $$ \begin{aligned} &\therefore \mathrm{p}(3)=\left(3^{3}-2 \times 3^{2}-5 \times 3+6\right)=(27-18-15+6)=0 \\...
is . Find the other zeros of the polynomial.$x=\frac{2}{3}$ is one of the zero of $3 x^{3}+16 x^{2}+15 x-18$ Now, we have $\mathrm{x}=\frac{2}{3}$ $\Rightarrow \mathrm{x}-\frac{2}{3}=0$ Now, we divide $3 x^{3}+16 x^{2}+15 x-18$ by...
and 
are the roots of the quadratic equation 
then find the value of a and 
.$a x^{2}+7 x+b=0$ Since, $x=\frac{2}{3}$ is the root of the above quadratic equation Hence, it will satisfy the above equation. => $a\left(\frac{2}{3}\right)^{2}+7\left(\frac{2}{3}\right)+b=0$...
and their product is 1 . Hence, find the zeroes of the polynomial.Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $\mathrm{f}(\mathrm{x})$. =>$(\alpha+\beta)=\frac{5}{2}$ and $\alpha \beta=1$ $\therefore...
Hence, find the zeroes of the polynomial.Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $\mathrm{f}(\mathrm{x})$. Then $(\alpha+\beta)=8$ and $\alpha \beta=12$ $\therefore f(x)=x^{2}-(\alpha+\beta) x+\alpha \beta$...
Verify the relation between the coefficients and the zeroes of the polynomial.Let $\alpha=2$ and $\beta=-6$ Sum of the zeroes, $(\alpha+\beta)=2+(-6)=-4$ Product of the zeroes, $\alpha \beta=2 \times(-6)=-12$ $\therefore$ Required polynomial $=\mathrm{x}^{2}-(\alpha+\beta)...
and verify the relation between the zeroes and the coefficients.$$ \begin{aligned} &3 x^{2}-x-4=0 \\ &\Rightarrow 3 x^{2}-4 x+3 x-4=0 \\ &\Rightarrow x(3 x-4)+1(3 x-4)=0 \\ &\Rightarrow(3 x-4)(x+1)=0 \\ &\Rightarrow(3 x-4) \text { or }(x+1)=0 \\ &\Rightarrow...
and verify the relation between the zeroes and the coefficients.f(u)=5u2+10u \mathrm{f}(\mathrm{u})=5 \mathrm{u}^{2}+10 \mathrm{u} It can be written as $5 \mathrm{u}(\mathrm{u}+2)$ ∴f(u)=0⇒5u=0 or u+2=0 \therefore \mathrm{f}(\mathrm{u})=0...
and verify the relation between the zeroes and the coefficients.$$ \begin{aligned} &\mathrm{f}(\mathrm{x})=8 \mathrm{x}^{2}-4 \\ &\text { It can be written as } 8 \mathrm{x}^{2}+0 \mathrm{x}-4 \\ &=4\left\{(\sqrt{2} x)^{2}-(1)^{2}\right\} \\ &=4(\sqrt{2}...
and verify the relation between the zeroes and the coefficients.$$ \begin{aligned} &f(x)=x^{2}-5 \\ &\text { It can be written as } x^{2}+0 x-5 . \\ &=\left(x^{2}-(\sqrt{5})^{2}\right) \\ &=(x+\sqrt{5})(x-\sqrt{5}) \\ &\therefore f(x)=0...
and verify the relation between the zeroes and the coefficients.$$ \begin{aligned} &4 x^{2}-4 x+1=0 \\ &\Rightarrow(2 x)^{2}-2(2 x)(1)+(1)^{2}=0 \end{aligned} $$ $$ \begin{aligned} &\Rightarrow(2 \mathrm{x}-1)^{2}=0 \quad\left[\because \mathrm{a}^{2}-2...
and verify the relation between its zeroes and coefficients.$$ \begin{aligned} &2 \sqrt{3} x^{2}-5 x+\sqrt{3} \\ &\Rightarrow 2 \sqrt{3} x^{2}-2 x-3 x+\sqrt{3} \\ &\Rightarrow 2 x(\sqrt{3} x-1)-\sqrt{3}(\sqrt{3} x-1)=0 \\ &\Rightarrow(\sqrt{3} x-1) \text {...
and verify the relationship between the zeroes and coefficients of the given polynomial.$f(x)=5 x^{2}-4-8 x$ $=5 x^{2}-8 x-4$ $=5 x^{2}-(10 x-2 x)-4$ $=5 x^{2}-10 x+2 x-4$ $=5 x(x-2)+2(x-2)$ $=(5 x+2)(x-2)$ $\therefore \mathrm{f}(\mathrm{x})=0 \Rightarrow(5...
and verify the relation between its zeroes and coefficients.$f(x)=4 x^{2}-4 x-3$ $=4 x^{2}-(6 x-2 x)-3$ $=4 x^{2}-6 x+2 x-3$ $=2 x(2 x-3)+1(2 x-3)$ $=(2 x+1)(2 x-3)$ $\therefore \mathrm{f}(\mathrm{x})=0 \Rightarrow(2 \mathrm{x}+1)(2 \mathrm{x}-3)=0$...
and verify the relation between its zeroes and coefficients.$f(x)=x^{2}+3 x-10$ $=x^{2}+5 x-2 x-10$ $=x(x+5)-2(x+5)$ $=(x-2)(x+5)$ $\therefore \mathrm{f}(\mathrm{x})=0 \Rightarrow(\mathrm{x}-2)(\mathrm{x}+5)=0$ $\Rightarrow x-2=0$ or $x+5=0$ $\Rightarrow...
and verify the relation between its zeroes and coefficients.$x^{2}-2 x-8=0$ $\Rightarrow \mathrm{x}^{2}-4 \mathrm{x}+2 \mathrm{x}-8=0$ $\Rightarrow x(x-4)+2(x-4)=0$ $\Rightarrow(x-4)(x+2)=0$ $\Rightarrow(x-4)=0$ or $(x+2)=0$ $\Rightarrow x=4$ or $x=-2$ Sum...
and verify the relation between its zeroes and coefficients.$x^{2}+7 x+12=0$ $\Rightarrow x^{2}+4 x+3 x+12=0$ $\Rightarrow x(x+4)+3(x+4)=0$ $\Rightarrow(x+4)(x+3)=0$ $\Rightarrow(x+4)=0$ or $(x+3)=0$ $\Rightarrow \mathrm{x}=-4$ or $\mathrm{x}=-3$ Sum of...
, if two of its zeroes are 
and 
Due to the fact that this is a polynomial equation of degree 4, there will be a total of four roots. $V(5 / 3)$ and $-\sqrt{(5 / 3)}$ are zeroes of polynomial $f(x)$ $\therefore(x-\sqrt{(} 5 /...
by a polynomial 
, the quotient and remainder were 
and 
, respectively. Find Solution: Given, Dividend, $p(x)=x^{3}-3 x^{2}+x+2$ Quotient $=x-2$ Remainder $=-2 x+4$ We must find the value of Divisor, $g(x)$ Concept: We already know that, Dividend = Divisor $\times$ Quotient...
Dividend p(x) and divisor g(x) are two polynomials, according to the division method, where g(x)=0. Then, using the formula below, we can get the value of the quotient q(x) and the remainder r(x);...
![\[\text { (i) } p(x)=x^{3}-3 x^{2}+5 x-3, g(x)=x^{2}-2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d4dcd3ee9779f4b84bed2bef3209dcb1_l3.png "Rendered by QuickLaTeX.com")
![\[\text { (ii) } p(x)=x^{4}-3 x^{2}+4 x+5, g(x)=x^{2}+1-x\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-baa8832c9a3a510eaa029756371b65a6_l3.png "Rendered by QuickLaTeX.com")
![\[\text { (iii) } p(x)=x^{4}-5 x+6, g(x)=2-x^{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-31d1cd4548678d9def575c9a8b85541f_l3.png "Rendered by QuickLaTeX.com")
Solution: Solution: i) Given, Dividend $$=p(x)=x^{3}-3 x^{2}+5 x-3$$ Divisor $$=g(x)=x^{2}-2$$ We use long division method: Therefore, upon division we get, Quotient $=x-3$ Remainder $$=7 x-9$$ ii)...
Solution: Concept: It is the zeros of a polynomial that correspond to the values of x that fulfil the equation y = f. (x). Here, f(x) is a function of x, and the zeros of the polynomial represent...
Solutions: Concept: When a polynomial is zeroed, the values of x that meet the equation y = f are considered zeros (x). There is a f(x) function, and the zeros of a polynomial are all of the values...
Solutions: The following is a graphical way for finding zeroes:- When solving any polynomial equation, the total number of zeroes equals the total number of times the curve meets the x-axis. i) The...