Question

Factorise 9(a+b)^2 + (a+b) -8 ​

Answer 1

Answer:

$(9a+9b-8)(a+b+1)$

Step-by-step explanation:

$9(a+b)^2 + (a+b)-8$

Expand $(a+b)^2$

$9(a^2+2ab+b^2)+a+b-8$

$9a^2+18ab+9b^2+a+b-8$

Now it is time to think.

Once

$9a^2=9a \cdot a$

$9b^2=9b \cdot b$

$18ab=9ab+9ab$

$-8=-8 \cdot 1$

We can get something in the format

$(9a+9b-8)(\text{some expression here})$

From the conclusions above,

$(9a+9b-8)(a+b+1)$