Test: Factorisation- 2 - Class 8 MCQ

Now, 12a² b + 15ab²

= (3 * 2 * 2 * a * a * b) + (3 * 5 * a * b * b)

= (3 * a * b) {(2 *2 * a) + (5 * b)}

= 3ab (4a + 5b)

C ) (x² +8x +16)
= (x² + 4x + 4x + 16)
= ( x² + 4x ) + (4x + 16 )
= x(x + 4 ) +4(x + 4 )
=(x + 4)(x + 4)
=(x + 4)² 

When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions. 

x² + 8x + 12 
two no whose product is 12 and sum is 8
ie. 2 and 6 so;
x²+2x+6x+12
x(x+2)+6(x+2)
(x+2)(x+6)

–18 xyz²/–3 xz   x and x gets cancelled,-18 gets divided by -3 .and z² gets divided by z so only one z remain in numerator So the quotient obtained is 6yz

We have  p4 - 81 = (p²)² - (9)² 

Now using a² - b² = (a + b)(a - b), we have

         (p²)² - (9)² = (p² + 9)(p² -9)

We can factorise p² - 9  further as

              p² - 9 = (p)² - (3)²

                       = (p + 3)(p - 3)

∴           p⁴ - 81 = (p + 3)(p - 3)(p² + 9)