How do you factor #4x^4-5x^2-9#?

1 Answer
May 26, 2018

Answer:

#(2x-3)(2x+3)(x+i)(x-i)#

Explanation:

#"using the a-c method to factor"#

#"the factors of the product "4xx-9=-36#

#"which sum to - 5 are + 4 and - 9"#

#"split the middle term using these factors"#

#4x^4+4x^2-9x^2-9larrcolor(blue)"factor by grouping"#

#=4x^2(x^2+1)-9(x^2+1)#

#"take out the "color(blue)"common factor "(x^2+1)#

#=(x^2+1)(4x^2-9)#

#4x^2-9" is a "color(blue)"difference of squares"#

#"which factors in general as"#

#•color(white)(x)a^2-b^2=(a-b)(a+b)#

#4x^2=(2x)^2rArra=2x#

#9=(3)^2rArrb=3#

#4x^2-9=(2x-3)(2x+3)#

#x^2+1" can also be factored using difference of squares"#

#=(x+i)(x-i)#

#4x^4-5x^2-9=(2x-3)(2x+3)(x+i)(x-i)#