Question

Factor the polyomial completely using any method 6b^4-17b^2-28
Medal given for answer

Answer 1

rewrite 17b^2 as -24b^2+7b^2 since -24*7=-168 and -24+7=-17 (the -168 comes from the product of ac=6*(-28)=-168)\[=6b^4-24b^2+7b^2-28\]factor by grouping (take GCF from first two terms and second two terms)\[=6b^2(b^2-4)+7(b^2-4)\]now factor out the binomial as a GCF\[=(b^2-4)(6b^2+7)\]the first binomial is a difference of two squares, to complete the factoring\[=(b-2)(b+2)(6b^2+7)\]