Question

9.
Factorised form of (x2 + 2x) whole square - 5(x2 + 2x) - 6
a) (x + 1)(x² + 2x - 6)
b) (x²+2x - 6)
c) (x - 1) whole square (x² + 2x-6)
d) (x - 1) whole square (x2 - 2x - 6)​

Answer 1

Step-by-step explanation:

Let f(x)=x

3

−2x

2

−5x+6

⇒f(1)=1

3

−21

2

−5×1+6=1−2−5+6=0

Using synthetic division, we have

x

2

−x−6=0 is the other factor

(x−1)(x

2

−x−6)=0

=(x−1)(x

2

−3x+2x−6)=0

=(x−1)(x(x−3)+2(x−3))=0

=(x−1)(x+2)(x−3) are the factors of f(x)

Answer 2

$\textsf{\large{\underline{Solution}:}}$

We have to factorise the given polynomial.

$\sf = {( {x}^{2} + 2x)}^{2} - 5( {x}^{2} + 2x ) - 6$

Let us assume that:

$\sf: \longmapsto u = {x}^{2} + 2x$

Therefore, the polynomial becomes:

$\sf = {u}^{2} - 5u - 6$

By splitting the middle term, we get:

$\sf = {u}^{2} - 6u + u - 6$

$\sf = u(u - 6) + 1(u - 6)$

$\sf = (u + 1)(u - 6)$

Substitute back u = x² + 2x, we get:

$\sf = ( {x}^{2} + 2x + 1)( {x}^{2} + 2x - 6)$

$\sf = ( {x}^{2} + x + x+ 1)( {x}^{2} + 2x - 6)$

$\sf = (x(x + 1)+ 1(x+ 1))( {x}^{2} + 2x - 6)$

$\sf = (x + 1)(x+ 1)( {x}^{2} + 2x - 6)$

$\sf = (x + 1)^{2} ( {x}^{2} + 2x - 6)$

Which is our required answer.

$\textsf{\large{\underline{Answer}:}}$

• The factorised form of the given polynomial is (x + 1)²(x² + 2x - 6)