Question

Factorise the expression 6k + 2 - 6kh - 2h

Answer 1

Answer:

$\bf { (2-2h)(3k + 1)}$

Step-by-step explanation:

As per the provided question, we have to factorise the given polynomial,

• $\rm { 6k +2 - 6kh - 2h}$

Here, it is necessary to re-arrange the given terms into groups, so the the group have a common factor.

$\longmapsto\rm { 6k +2 - 6kh - 2h}$

Here, think that (6k + 2) is 2(3k + 1) and (-6kh - 2h) is -2h(3k + 1) and so get the common factor.

Basically, we have to work out mentally as to which arrangement would lead to groups yielding a common binomial factor.

$\longmapsto\rm { 2(3k +1) - 2h(3k + 1)}$

Now, take (3k + 1) as common.

$\longmapsto\bf { (2-2h)(3k + 1)}$

Hence factorized!

More Information :

• Factorisation is the reverse process of finding the product.
• To factorise, we take out the common monomial factor.

Some identities used in factorisation :

• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab
• a² - b² = (a + b)(a - b)

Answer 2

$\large\sf\red{\underline{Question}}:$
• Factorise the expression 6k + 2 - 6kh - 2h.
$\\\large\sf\red{\underline{Solution}}:$
$\longmapsto\sf 6k+2-6kh-2h \\\\\mapsto \textbf{Taking 2 common from expression}\\\\\implies 2[3k+1-3kh-h] \\\\\mapsto\textbf{Further taking like terms common} \\\\\implies 2[1(3k+1)-h(3k+1)] \\\\\implies 2[(3k+1)(1-h)] \\\\\implies\pink{ 2(3k+1)(1-h)}$

$\large\sf \underline{Therefore},$
➩ 2(3k+1)(1-h) is answer.