Problem 87 - Algebra and Trigonometry

Question

Factor completely, or state that the polynomial is prime. $$9 b^{2} x-16 y-16 x+9 b^{2} y$$

Step-by-Step Solution

Answer

The completely factored form of the given polynomial \(9 b^{2} x-16 y-16 x+9 b^{2} y\) is \((x + y)(3b-4)(3b+4)\).

Step 1: Rearrange the Polynomial

Rearrange the polynomial so that the like terms are together: \(9 b^{2} x - 16x + 9b^{2}y - 16y\)

Step 2: Factor by Grouping

Factor the polynomial in groups:\((9 b^{2} x - 16x) + (9b^{2}y - 16y) = x(9b^{2} - 16) + y(9b^{2} - 16)\)

Step 3: Common Factor

Now you can see that there exists a common factor \((9b^{2} - 16)\) in both terms, factor out the common factor: \((x + y)(9b^{2} - 16)\)

Step 4: Factor the Difference of Squares

Now, one more factorization step is left. The term \((9b^{2} - 16)\) is a difference of squares, so it can be factorized further: \((x + y)(3b-4)(3b+4)\)