Lesson 2: Multiplying Polynomials

Property

To multiply $(x+3)(x+7)$, you distribute the second binomial, $(x+7)$, to each term of the first binomial.This gives $x(x+7) + 3(x+7)$.
Then, you distribute again to get $x^2 + 7x + 3x + 21$.
Finally, combine like terms to get $x^2 + 10x + 21$.
Notice that you multiplied the two terms of the first binomial by the two terms of the second binomial, resulting in four multiplications.

Examples

• To multiply $(a+4)(a+6)$, distribute $(a+6)$: $a(a+6) + 4(a+6) = a^2 + 6a + 4a + 24$, which simplifies to $a^2 + 10a + 24$.
• For $(2x+1)(x-3)$, distribute $(x-3)$: $2x(x-3) + 1(x-3) = 2x^2 - 6x + x - 3$, which simplifies to $2x^2 - 5x - 3$.
• To multiply $(y-5)(z+2)$, distribute $(z+2)$: $y(z+2) - 5(z+2) = yz + 2y - 5z - 10$. There are no like terms to combine.

Explanation

This method breaks down the problem into smaller, familiar steps. You take the first term of the first binomial and multiply it by the entire second binomial, then do the same with the second term. It guarantees every piece gets multiplied.

Property

The horizontal method multiplies binomials by writing them side-by-side and systematically distributing each term from the first binomial to every term in the second binomial: $(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd$

Examples

Property

Another method that works for all polynomials is the Vertical Method.
It is very much like the method you use to multiply whole numbers.
You line up the binomials vertically, multiply the top binomial by each term of the bottom binomial to get partial products, and then add the partial products, making sure to align like terms in columns.

Examples

• To multiply $(x+5)(x-2)$ vertically, first multiply $-2(x+5)$ to get $-2x-10$. Then multiply $x(x+5)$ to get $x^2+5x$. Add the partial products: $(x^2+5x) + (-2x-10) = x^2+3x-10$.
• To multiply $(3a-4)(a+1)$ vertically, first multiply $1(3a-4)$ to get $3a-4$. Then multiply $a(3a-4)$ to get $3a^2-4a$. Add the partial products: $(3a^2-4a) + (3a-4) = 3a^2-a-4$.
• To multiply $(2y+3)(2y-3)$ vertically, first multiply $-3(2y+3)$ to get $-6y-9$. Then multiply $2y(2y+3)$ to get $4y^2+6y$. Add the partial products: $(4y^2+6y) + (-6y-9) = 4y^2-9$.

Explanation

This is like old-school multiplication with numbers, but for polynomials! You stack the binomials, multiply in parts, and add the results. It's a great way to keep your terms organized, especially with more complex polynomials.