Question

if one of the zeros of cubic polynomial

$x ^{3} + a {x}^{2} + bx + c$
is -1 then the product of other two zeros is

a)) b - a + 1
b)) b - a - 1
c)) a - b + 1
d)) a - b - 1


ANSWER SOON PLEASE ​

Answer 1

AnswEr :

$\sf{Given\:Polynomial : f(x) = x^3 + ax^2 + bx + c}\\\\\textsf{-1 is a zero of the given polynomial.}\\\\ \textsf{Let \alpha and \beta be other two zeroes of the polynomial.}$

$\rule{100}{2}$

$\longrightarrow \tt Sum \:of \:Zeroes = - \dfrac{b}{a} \\ \\ \\\longrightarrow \tt - 1 + \alpha + \beta = - \dfrac{a}{1} \\ \\ \\\longrightarrow \tt - 1 + \alpha + \beta = - a \\ \\ \\\longrightarrow \tt \alpha + \beta = 1 - a \qquad \dfrac{\quad}{}eq.(1)$

$\rule{200}{1}$

$\longrightarrow \tt Sum \:of \:Product \:of \:Zeroes = \dfrac{c}{a} \\ \\ \\\longrightarrow \tt ( - 1 \times \alpha) + ( \alpha \times \beta) + ( \beta \times - 1) = \dfrac{b}{1} \\ \\ \\ \longrightarrow \tt - \alpha + \alpha \beta - \beta = b \\ \\ \\\longrightarrow \tt \alpha \beta = b + (\alpha +\beta) \\ \\ \\\qquad \sf\star \:From\:eq.(1):[(\alpha + \beta) = 1-a] \\\\\\\longrightarrow \large\red{\tt\alpha\beta = b - a + 1}$

⠀⠀

$\therefore\boxed{\textsf{Product of Zero will be} \:\boxed{ \sf(a) \:b - a +1}}$

Answer 2

❏ Question:-

if one of the zeros of cubic polynomial $x ^{3} + a {x}^{2} + bx + c$ is -1

then the product of other two zeros is

a)) b - a + 1

b)) b - a - 1

c)) a - b + 1

d)) a - b - 1

❏ Solution:-

Now, one zero of the cubic polynomial is = -1

let, it $\bf \alpha= -1$

Now we have to find $\bf \beta\gamma$ in respect of a and b .

Now, comparing the equation ($\bf x ^{3} + a {x}^{2} + bx + c$)with (Px³+Qx²+Rx+S) is,

➝ P = 1

➝ Q = a

➝ R = b

➝ S = c

Now,

$\sf\longrightarrow\bf \alpha+\beta+\gamma=\frac{-Q}{P}$

$\sf\longrightarrow\bf (-1)+\beta+\gamma=\frac{-a}{1}$

$\sf\longrightarrow\bf \beta+\gamma=1-a$

Again,

$\sf\longrightarrow \bf\alpha\beta+\beta\gamma+\gamma \alpha=\frac{R}{P}$

$\sf\longrightarrow \bf (-1)\beta+\beta\gamma+\gamma (-1)=\frac{b}{1}$

$\sf\longrightarrow \bf\beta\gamma =b+(\beta+\gamma)$

$\sf\longrightarrow \bf\beta\gamma =b+(1-a)$

$\sf\longrightarrow\bf \boxed{\large{\red{\beta\gamma =b-a+1}}}$

Hence, Product of other two zeroes is (b-a+1)

Ans) Option (a)➝ (b-a+1).

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❏ Used Concept:-

✦ For a cubic polynomial :-

➝ F(x)=Px³+Qx²+Rx+S

if $\alpha\:,\beta\:,\gamma$ are the three roots of the cubic polynomial F(x).

Then,

$\sf\implies \bf\alpha+\beta+\gamma=\frac{-Q}{P}$

$\sf\implies \bf\alpha\beta+\beta\gamma+\gamma \alpha=\frac{R}{P}$

$\sf\implies \bf\alpha\beta\gamma=\frac{-S}{P}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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