Question

If α and β are the zeroes of the polynomial x²+3x-2(x+7) , then the value of (α + 1)(β+1) is

Answer 1

Given Polynomial:

$\sf {x}^{2} + 3x - 2(x + 7) \\ \sf \longrightarrow{x}^{2} + 3x - 2x - 14 \\ \sf \longrightarrow{x}^{2} + x - 14 \\ \sf \longrightarrow x^2 + x + (-14)$

It is now in the form of $\sf ax^2 + bx + c$

So,

$\sf \alpha + \beta = \frac{ - b}{a} = \frac{ - 1}{1} \\ \sf \longrightarrow\alpha + \beta = - 1\quad\quad\dots(1)$

And,

$\sf \alpha \beta = \frac{c}{a} = \frac{ - 14}{1} \\ \sf \longrightarrow \alpha \beta = - 14\quad\quad\dots(2)$

To find:

$\sf ( \alpha + 1)( \beta + 1) \\ \sf \longrightarrow \alpha \beta \: + \alpha + \beta + 1 \\ \sf \longrightarrow \alpha \beta + ( \alpha + \beta ) + 1 \\ \sf{From\:\: (1)\:\: and\:\: (2),} \\ \sf \longrightarrow - 14 + ( - 1) + 1 \\ \sf \longrightarrow - 14 - 1 + 1 \\ \longrightarrow - 14$

Hence, the answer is,

$\longrightarrow \underline{\underline{\sf{\green{(\alpha+1)(\beta+1) = -14}}}}$