Question

factorize it question 23 fast please

Answer 1

y^4 + 4y^2 - 32

= y^4 + 8y^2 - 4y^2 - 32

= y^2 ( y^2 + 8 ) - 4 ( y^2 + 8 )

= ( y^2 + 8 )( y^2 - 4 )

= ( y^2 + 8 )( y - 2 )( y + 2 )

Answer 2

hiii!!!

here's ur answer...

we will solve this by splitting method...

given y⁴ + 4y - 32

first we will find the multiples of 32 :-

→ 1 and 32

→ 2 and 16

→ 4 and 8

we have to find which of these pairs can be subtracted or added to make it 4.

only pair of 4 and 8. if we subtract 4 from 8, we get 4.

now,

next step..

= y⁴ + ( 8y² - 4y² ) - 32

= y⁴ + 8y² - 4y² - 32

now, divide it into two pairs. one is y⁴ + 8y² and other is 4y² - 32.

y⁴ = y × y × y × y

8y² = 2 × 2 × 2 × y × y

CF = y²

4y² = 2 × 2 × y × y

32 = 2 × 2 × 2 × 2 × 2

CF = 4

next,

y⁴ + 8y² = y² ( y² + 8 )

4y² - 32 = 4 ( y² + 8 )

= ( y² + 8 ) ( y² - 4 )

sorry if my answer is wrong...

:(