Lesson 2: Multiplying Polynomials

Property

To multiply two monomials, rearrange the factors to group together the numerical coefficients and the powers of each base.
Then, multiply the coefficients and use the first law of exponents for the variable factors.

Examples

• To multiply $(3a^2b)(4a^3b^4)$, we group and multiply: $(3 \cdot 4)(a^2 \cdot a^3)(b \cdot b^4) = 12a^5b^5$.

• The product of $(-6x^3y^2)(2x^5y)$ is found by multiplying coefficients and adding exponents of like bases: $(-6 \cdot 2)(x^3 \cdot x^5)(y^2 \cdot y) = -12x^8y^3$.

Property

When multiplying terms with the same base, add the exponents: $x^a \cdot x^b = x^{a+b}$

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.