Question

factorise 12x²+6root3x-4​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\: {12x}^{2} + 6 \sqrt{3}x - 4$

We use here the concept of

Splitting of middle terms :-

☆ In order to factorize  ax² + bx + c we have to find numbers p and q such that p + q = b and pq = ac.

☆ After finding p and q, we split the middle term in the quadratic as px + qx and get desired factors by grouping the terms.

❥ Calculation :-

Here,

We have to find the values of p and q such that

$\rm :\longmapsto\:p + q = 6 \sqrt{2} \: \: and \: pq = - 48$

Now,

$\rm :\longmapsto\:pq = - 48 \: can \: be \: factorised \: as$

$\rm :\longmapsto\:pq = - 16 \times 3$

$\rm :\longmapsto\:pq = - 8 \times 2 \times \times \sqrt{3} \times \sqrt{3}$

$\rm :\longmapsto\:pq = - (8 \sqrt{3}) \times (2 \sqrt{3})$

So that,

$\rm :\longmapsto\:p + q = 8 \sqrt{3} - 2 \sqrt{3}$

Hence,

$\rm :\longmapsto\: {12x}^{2} + 6 \sqrt{3}x - 4$

can be rewritten as

$\rm :\longmapsto\: {12x}^{2} + 8 \sqrt{3}x - 2 \sqrt{3}x - 4$

$\rm :\longmapsto\: {4 \times \sqrt{3} \times \sqrt{3} x}^{2} + 8 \sqrt{3}x - 2 \sqrt{3}x - 4$

$\rm :\longmapsto\:4 \sqrt{3}x( \sqrt{3}x + 2) - 2( \sqrt{3}x + 2)$

$\rm :\longmapsto\:( \sqrt{3}x + 2)(4 \sqrt{3}x - 2)$

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)