Question

@iambatman What is the completely factored form of d^4 - 8d^2 + 16?

(d2 + 4)(d2 - 4)
(d2 - 4)(d2 - 4)
(d2 + 4)(d + 2)(d - 2)
(d + 2)(d - 2)(d + 2)(d - 2)

Answer 1

i got d^2(d^2 - 8 + 16/d^2)

Answer 2

you can not factor out d^2 because there is no d^2 in 16

Answer 3

hint: \(\large {
d^4 - 8d^2 + 16\implies
\begin{array}{cccllll}
({\color{brown}{ d^2}})^2&-8{\color{brown}{ d^2}}&+16\\
&-4-4&-4\cdot -4
\end{array}
}\)

Answer 4

d^4 - 8d^2 + 16
= (d^2 - 4)(d^2 - 4)

now you factor d^2 - 4

- the difference of 2 squares
do you recall what these fact to?

Answer 5

a^2 - b^2 = (a + b)(a - b)

Answer 6

OHHH so its B

Answer 7

no - it will factor further..

Answer 8

close but you have to factor further
the (d^2-4) terms can be factored like @welshfella said

Answer 9

ok how do i do that then

Answer 10

oh wait just the symbols so A ?

Answer 11

take square root of d^2 and square root of 4
- these are the a and b in the formula i posted

Answer 12

ok.... d and 2