How do you factor #4x^4-5x^2-9#?
1 Answer
May 26, 2018
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)#