Question

Y2-5y-6
Factorise using factor theorem

Answer 1

Answer:

$y^{2} - 5y - 6$

$= y^{2} - 6y + y - 6$

$= y(y - 6) +1(y - 6)$

$= (y - 6)( y + 1)$

Answer 2

By definition of FACTOR theorum,

If ' a ' is a root ( or zero ) of a polynomial, then ( y - a ) is a factor of that polynomial.

Let p(y) be that polynomial.

Therefore, p(a) = 0

Method used for using factor theorum :— Hit & Trial

So, we'll find that ( 6 ) and ( - 1 ) are the zeroes of the polynomial.

let's ☑ check :

$\tt \quad p(6) = {(6)}^{2} - 5(6) - 6 \\ \\ \qquad \quad = 36 - 30 - 6 \\ \\ \qquad \quad = 0$
And,

$\tt \quad p( - 1) = {( - 1)}^{2} - 5( - 1) - 6 \\ \\ \qquad \quad \: \: \: \: = 1 + 5 - 6 \\ \\ \qquad \quad \: \: \: \: = 0$
Hence, by FACTOR THEORUM, it's clear that ( 6 ) and ( - 1 ) are the zeroes of the polynomial $\bf {y}^{2} - 5y - 6$ .