Find the product (5x^2 - 3x)(2x^2 + 6x - 9), writing the multiplication horizontally.

Step-by-step solution

1. 1

   Set up the distribution. A binomial times a trinomial produces $2\times 3=6$ partial products. Take each term of the first factor in turn and multiply it by all three terms of the second factor, keeping the signs attached to the terms.

2. 2

   Distribute the first term $5x^2$. $5x^2\cdot 2x^2=10x^4$, $5x^2\cdot 6x=30x^3$, and $5x^2\cdot(-9)=-45x^2$. Exponents add because $x^a\cdot x^b=x^{a+b}$.

3. 3

   Distribute the second term $-3x$. $-3x\cdot 2x^2=-6x^3$, $-3x\cdot 6x=-18x^2$, and $-3x\cdot(-9)=+27x$. The last sign flips because a negative times a negative is positive.

4. 4

   Write all six partial products in one line. $10x^4+30x^3-45x^2-6x^3-18x^2+27x$.

5. 5

   Combine like terms by degree. Degree 4: $10x^4$ stands alone. Degree 3: $30x^3-6x^3=24x^3$. Degree 2: $-45x^2-18x^2=-63x^2$. Degree 1: $27x$. There is no constant term because every term of the first factor carries an $x$.

6. 6

   Check with a test value. At $x=1$ the factors are $5-3=2$ and $2+6-9=-1$, so the product should be $-2$; the expansion gives $10+24-63+27=-2$. At $x=2$: $(20-6)(8+12-9)=14\cdot 11=154$, and $160+192-252+54=154$.