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

( d - 2 )*( d + 2 )*( d - 2 )*( d + 2 )

Step-by-step explanation:

Given:-

- The quartic expression is given as follows with the independent variable as "d".

                                   d^4 - 8d^2 + 16

Find:-

What is the completely factored form of the given quartic expression.

Solution:-

- The expression can be re-written in a quadratic formulation. " Reducible quadratic ". We will make substitution y = d^2.

                                   (y)^2 - 8(y) + 16

- We can use the substituted expression and find the factors. So possible factor values of (16) are:

                                    8*2 = 16 , 10 , 6

                                    -4*-4 = 16 , -8, 0

-  So we will use the pair ( -4 , -4 ) as the common factor:

                                    ( y - 4 )*( y - 4 ) =  (y)^2 - 8(y) + 16

- Now use back-substitution of y = d^2, we have:

                                   ( d^2 - 4 ) * ( d^2 - 4 )

- We can use the perfect square to express the form as:

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

                                   ( d - 2 )*( d + 2 )*( d - 2 )*( d + 2 )  

answer: asdfghjkl,mnbv

step-by-step explanation:

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.