Question

Factor completely: x2 - 8x + 16

Answer 1

please explain D;

Answer 2

x^2-8x+16

x^2-4x-4x+16 ... see note below

(x^2-4x)+(-4x+16)

x(x-4)-4(x-4)

(x-4)(x-4)

(x-4)^2


So x^2-8x+16 completely factors to (x-4)^2


Note: we want to find two numbers that multiply to 16 and add to -8. These two numbers are -4 and -4. So we break -8x into -4x-4x and then factor by grouping.

Answer 3

now we look at the first term and the last term
so x^2 and 16
we need to brackets
( )( )
(x )(x ) we can see the x^2 term comes from x times x
(x-4 )(x-4) why did i put -4 in the first and last term because those two numbers multiply to equal positive 16(referring to the last term) and add togeth -4-4 to equal the middle term which is -8.

Answer 4

So the answer is (x-4) (x-4)?

Answer 5

yes or you can write the answer as (x-4)^2 as they are the same answer.