Question

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

Answer 1

Solution:

1) Rewrite it in the form {a}^{2}-2ab+{b}^{2}, where a={d}^{2} and b=4
{({d}^{2})}^{2}-2({d}^{2})(4)+{4}^{2}

2) Use Square of Difference: {(a-b)}^{2}={a}^{2}-2ab+{b}^{2}
{({d}^{2}-4)}^{2}

3) Rewrite {d}^{2}-4​ in the form {a}^{2}-{b}^{2}​​ , where a=d and b=2
{({d}^{2}-{2}^{2})}^{2}
​​
4) Use Difference of Squares: {a}^{2}-{b}^{2}=(a+b)(a-b)
​{((d+2)(d-2))}^{2}

5) Use Multiplication Distributive Property: {(xy)}^{a}={x}^{a}{y}^{a}
{(d+2)}^{2}{(d-2)}^{2}

Done!

Answer 2

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

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 )