How do you factor #4m^4 - 37m^2 + 9#?

2 Answers
Nov 13, 2016

Answer:

#(4m^2 - 1)(m^2 - 9)#

Explanation:

Factoring this polynomial gives:

#(4m^2 - 1)(m^2 - 9)#

Nov 18, 2016

Answer:

#4m^4-37m^2+9 = (2m-1)(2m+1)(m-3)(m+3)#

Explanation:

The difference of squares identity can be written:

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

We will use this a couple of times, but first note that:

#4m^4-37m^2+9 = 4(m^2)^2-37(m^2)+9#

So we can treat this quartic as a quadratic in #m^2#

Factor using an AC method:

Look for a pair of factors of #AC = 4*9 = 36# with sum #B=37#

The pair #36, 1# works.

Use this pair to split the middle term and factor by grouping:

#4(m^2)^2-37(m^2)+9 = 4(m^2)^2-36(m^2)-(m^2)+9#

#color(white)(4(m^2)^2-37(m^2)+9) = (4(m^2)^2-36(m^2))-((m^2)-9)#

#color(white)(4(m^2)^2-37(m^2)+9) = 4m^2(m^2-9)-1(m^2-9)#

#color(white)(4(m^2)^2-37(m^2)+9) = (4m^2-1)(m^2-9)#

#color(white)(4(m^2)^2-37(m^2)+9) = ((2m)^2-1^2)(m^2-3^2)#

#color(white)(4(m^2)^2-37(m^2)+9) = (2m-1)(2m+1)(m-3)(m+3)#