Question

What is the completely factored form of d4 − 8d2 + 16?

Answer 1

are you able to factor z^2 - 8z + 16 ?

Answer 2

What is z? o.o

Answer 3

Uhhhhhh I'm hella lost lol. ;^;

Answer 4

I let z = d^2

so z^2 = (d^2)^2 = d^4

Answer 5

(Z-4)^2

Answer 6

d^4 - 8d^2 + 16 will turn into z^2 - 8z + 16

Answer 7

To cancel out ^2? Er I just started this module today. Spare me .-.

Answer 8

have you factored quadratics before?

Answer 9

Yeah I did a couple. This one confuses me though because it starts with a variable.

Answer 10

what types did you factor before?

Answer 11

Like 3x^3+12x^2+18x

Answer 12

how did you factor that

Answer 13

Welllllll I factor out the GCF. Then I factor out the variables.

3x^3+12x^2+18x

3(x^3+4x^2+6x)

3x(x^2+4x+6)

Answer 14

http://www.purplemath.com/modules/factquad.htm

Answer 15

Correct. To factor something like z^2 - 8z + 16, we need to follow these steps


step 1) find two numbers that multiply to 16 (last term) and add to -8 (middle term)
these two numbers are -4 and -4
-4 plus -4 = -8
-4 times -4 = +16

step 2) take the two numbers and use them to write out the factorization
z^2 - 8z + 16 = (z-4)*(z-4)