Question

Factorise: – 2(5x-12) – XP (3 – X).
Bta do bhut merh bani ho gi agar kisi ne bta diya tohhh???

Answer 1

Given:-

• -2(5x - 12) - x²(3 - x)

We have to factorise it.

Answer:-

-2(5x - 12) - x²(3 - x)

Expanding it,

→ -10x + 24 - 3x² + x³

Rewriting it,

→ x³ - 3x² - 10x + 24

Now, have to find a zero of the above polynomial. There's no "direct" way of doing it, we have to find a value through hit and trial. Usually the value comes within -2, -1, +1 or +2

Putting x = 2

→ 2³ - 3(2²) - 10x + 24

→ 8 - 12 - 20 + 24

→ 0

So, 2 is a zero of the above polynomial.

Now, through Factor Theorem, we can say that,

(x - 2) is a factor of the above polynomial

Till now, we have simplified till,

→ x³ - 3x² - 10x + 24

→ (x - 2) (x - 2) (x - 2)

In these blank spaces, we have to write the suitable variable/constant

The first term in the polynomial is x³. So, in the first blank space we can put x² in order to get x³.

→ x²(x - 2) (x - 2) (x - 2)

As a result, we have got: x³ - 2x² .....

Next after this, we want -3x². So we need (-x²) so that, -2x² - x² = -3x². So, putting (-x) in the second blank space,

→ x²(x - 2) -x(x - 2) (x - 2)

As a result we have got: x³ - 2x² - x² + 2x ......

= x³ - 3x² + 2x ......

Next after this, we want (-10x + 24). So putting (-12) in third blank space such that, -12(x - 2) = -12x + 24 and finally getting: +2x -12x + 24 = -10x + 24

→ x²(x - 2) -x(x - 2) -12(x - 2)

(and this is equal to x³ - 3x² - 10x + 24)

Taking (x - 2) as common,

→ (x - 2)(x² - x - 12)

Splitting the middle term of (x² - x - 12),

→ (x - 2)(x² - 4x + 3x - 12)

→ ( x - 2 ){ x(x - 4) + 3(x - 4) }

Taking (x - 4) as common,

→ (x - 2)(x - 4)(x + 3) Ans.