College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.5 - Factoring Polynomials - R.5 Exercises: 119

Answer

$(3x-5)(9x^2-3x+7)$

Work Step by Step

With $27=3^3$, the given expression is equivalent to: $=(3x-2)^3-3^3$ RECALL: The difference of two cubes $a^3-b^3$ can be factored using the formula: $a^3-b^3=(a-b)(a^2+ab+b^2)$ Factor the given difference of two cubes using the formula above with $a=3x-2$ and $b=3$ to obtain: $(3x-2)^3-3^3 \\=[(3x-2)-3][(3x-2)^2+(3x-2)(3)+3^2] \\=(3x-2-3)[(9x^2-2(3x)(2)+2^2)+(9x-6)+9] \\=(3x-5)[9x^2-12x+4+9x-6+9]$ Combine like terms to obtain: $=(3x-5)[9x^2+(-12x+9x)+(4-6+9)] \\=(3x-5)(9x^2-3x+7)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.