Question

4x^8-61x^4+225 ... the answer is (2x^2+5)(2x^2-5)(x^2+3)(x^2-3) but i know to know how to get there

Answer 1

\[4x^8-61x^4+225\]
\[\rightarrow (2x^2+5) (2x^2-5)(x^2+3)(x^2-3) \]
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Answer 2

4x^(8)-61x^(4)+225

For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (900) that add up to b (-61).In this problem -(25)/(4)*-9=(225)/(4) (which is (c)/(a)) and -(25)/(4)-9=-(61)/(4) (which is ((b)/(a)) , so insert -(25)/(4) as the right hand term of one factor and -9 as the right-hand term of the other factor.
(x^(4)-(25)/(4))(x^(4)-9)

Remove the fraction by multiplying the first term of the factor by the denominator of the second term.
(4x^(4)-25)(x^(4)-9)

The binomial can be factored using the difference of squares formula, because both terms are perfect squares. The difference of squares formula is a^(2)-b^(2)=(a-b)(a+b).
(2x^(2)-5)(2x^(2)+5)(x^(4)-9)


=(2x^(2)-5)(2x^(2)+5)(x^(2)+3)(x^(2)-3)