Question

if alpha and beta are the zeros of the polynomial ax2+bx+c find the value of (alpha+1) (beta+1)

Solve it the clearest for brainliest

Answer 1

ax²+bx+c is the polynomial

let

$\alpha \\ \beta \\ be \: the \: roots$

$\alpha + \beta = \frac{ - b}{a}$

$\alpha \beta = \frac{c}{a}$

now,

$( \alpha + 1)( \beta + 1) = \\ \alpha \beta + \alpha + \beta + 1$

$\frac{c}{a} + ( \frac{ - b}{a} + 1)$

$\frac{c - b}{a} + 1$

take lcm

$\frac{a + c - b}{ a}$

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Answer 2

Given

A polynomial ax²+bx+c

To Find

$(\alpha+1)(1+\beta)$

Solution

If the zeroes are $(\alpha)&(\beta)$

We know that

$\alpha+\beta=\frac{-b}{a}$

&

$\alpha\times\beta=\frac{c}{a}$

Putting values in it

$(\alpha+1)(1+\beta)$

Solving it

$\imples(\alpha\times\beta+\alpha+\beta+1)$

=>$\frac{-b}{a}+\frac{c}{a}$+1

=>$\frac{-b+c+a}{a}$