Question

Factorization of 4x^2 +9 -12x -a^2 -b^2 +2ab

Answer 1

Answer:

4x^2 + 9 – 12x – a^2 – b^2 + 2ab

= (4x^2 – 12x + 9) – (a^2 + b^2 – 2ab)

= (2x – 3)^2 – (a – b)^2

= [(2x – 3) + (a – b)] [(2x – 3) – (a – b)]

= (2x – 3 + a – b)(2x – 3 – a + b)

Answer 2

The factors of $\bf 4 {x}^{2} + 9 - 12x - {a}^{2} - {b}^{2} + 2ab$ are $\bf \red{(2x -3 + a - b)(2x - 3 - a + b)}$.

Given:

• $4 {x}^{2} + 9 - 12x - {a}^{2} - {b}^{2} + 2ab$

To find:

• Factorise the polynomial.

Solution:

Identity to be used:

1. $\bf ( {a - b)}^{2} = {a}^{2} - 2ab + {b}^{2}$
2. $\bf {p}^{2} - {q}^{2} = (p + q)(p - q)$

Step 1:

Make whole square of first three terms.

Rewrite the terms.

$\red{{(2x)}^{2} + (3) ^{2} - 2(2x)(3)} - {a}^{2} - {b}^{2} + 2ab$

using Identity 1 we can write it as

$( {2x - 3)}^{2} - {a}^{2} - {b}^{2} + 2ab$

take (-1) common from last three terms and convert these into whole square.

$( {2x - 3)}^{2} - ( {a}^{2} + {b}^{2} - 2ab)$

or

$\bf ( {2x - 3)}^{2} - {(a - b)}^{2}$

Step 2:

Compare the expression with identity 2.

It is clear that

$\bf p = 2x - 3$

and

$\bf q = a - b$

So, apply Identity 2 to factorise the polynomial.

$( {2x - 3)}^{2} - {(a - b)}^{2} = (2x - 3 + a - b)(2x - 3 - a + b)$

Thus,

The factors of $\bf 4 {x}^{2} + 9 - 12x - {a}^{2} - {b}^{2} + 2ab$ are $\bf (2x -3 + a - b)(2x - 3 - a + b)$.

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