Find the product (6x² − 7)(5x² − 6).

Step-by-step solution

1. 1

   Treat x² as one block. Substituting $u = x^2$ turns the problem into $(6u - 7)(5u - 6)$, an ordinary binomial product; this keeps the exponent bookkeeping honest.

2. 2

   Multiply the first terms. $6u \cdot 5u = 30u^2$, which is $30x^4$ once $u = x^2$ is restored.

3. 3

   Multiply the outer and inner pairs. $6u \cdot (-6) = -36u$ and $(-7) \cdot 5u = -35u$. Both are negative because each product mixes one positive and one negative term.

4. 4

   Multiply the last terms. $(-7)(-6) = 42$: two negatives give a positive constant.

5. 5

   Combine and translate back. $-36u - 35u = -71u$, so the product is $30u^2 - 71u + 42 = 30x^4 - 71x^2 + 42$.

6. 6

   Check at x = 1. The factors give $(6-7)(5-6) = 1$, and the expansion gives $30 - 71 + 42 = 1$.