Question

factorise the following​

Answer 1

Answer:

${x}^{3} + {x}^{2} \: + a + a + 2b \: + 2b \: \\ {x}^{3} + {x}^{2} \: + a + a \: + 2b \: + 2b \\ {x}^{5} + 2a \: + 4b \:$

Hope it helps you

Step-by-step explanation:

3+2 is 5 that's why I take five there

And as we know if there is no symbol there is always +

Answer 2

Answer:

Answers below:

Step-by-step explanation:

(b)

$(x+y)(2x+3y)-(x+y)(x+1)$

as you can see here, (x+y) is common in both lhs and rhs,

so taking (x+y) common,

$(x+y)[(2x+3y) - (x+1)]\\\\(x+y)[2x+3y-x-1]\\\\(x+y)[x+3y-1]\\\\\\Ans:(x+y)(x+3y-1)$

(c)

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

now expand this equation:

$x^{3} (a+2b)+x^{2} (a+2b)\\\\=ax^{3} +2bx^{3}+ax^{2} +2bx^{2}$

now regroup the terms with most similarities from left to right:

$ax^{3} +ax^{2} +2bx^{3} +2bx^{2}$

factorize the first 2 terms and last 2 terms separately:

$ax^{2} (x+1) +2bx^{2} (x+1)$

here you can see that (x+1) is common in both lhs and rhs.

so, use (x+1) as a common and the following should be your answer:

$(x+1)(ax^{2} +2bx^{2} )$  

Thanks.

Hope it helped.