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6.03 Multiplying special products

Lesson

Concept summary

Special products are special cases of products of polynomials. With special products, we can multiply two polynomials without using the distributive property.

For binomials, we have the following special binomial products:

Square of a binomial (Sum)

\left(a + b\right)^{2} = a^{2} + 2 a b + b^{2}

Square of a binomial (Difference)

\left(a - b\right)^{2} = a^{2} - 2 a b + b^{2}

Product of a sum and difference

\left(a+b\right)\left(a-b\right) = a^{2} - b^{2}

This is the identity of the difference of two squares.

Note: \left(a + b\right)^{2} \neq a^{2} + b^{2} and \left(a - b\right)^{2} \neq a^{2} - b^{2}

Worked examples

Example 1

Multiply and simplify: \left(x - 4\right)^{2}

Approach

We check first whether \left(x - 4\right)^{2} is a special binomial product and identify its form.

Solution

Since \left(x - 4\right)^{2} is a square of a binomial in the form \left(a - b\right)^{2}, we use the formula and simplify the expression:

\displaystyle \left(a - b\right)^{2}\displaystyle =\displaystyle a^{2} - 2 a b + b^{2}Formula for the square of a binomial
\displaystyle \left(x - 4\right)^{2}\displaystyle =\displaystyle x^{2} - 2\left(x\right)\left(4\right) + 4^{2}Substitute the given values
\displaystyle =\displaystyle x^{2} - 8 x + 16Simplify

Example 2

Multiply and simplify: \left(x + 4\right)\left(x - 4\right)

Approach

We check first whether \left(x + 4\right)\left(x - 4\right) is a special binomial product and identify its form.

Solution

Since \left(x + 4\right)\left(x - 4\right) is a product of a sum and difference, we use the formula and simplify the expression:

\displaystyle \left(a+b\right)\left(a-b\right)\displaystyle =\displaystyle a^{2} - b^{2}Formula for the product of a sum and difference
\displaystyle \left(x+4\right)\left(x-4\right)\displaystyle =\displaystyle x^{2} - 4^{2}Substitute the given values
\displaystyle =\displaystyle x^{2} - 16Simplify

Example 3

Multiply and simplify: \left(2x + 5\right)\left(2x - 5\right)

Approach

We check first whether \left(2x + 5\right)\left(2x - 5\right) is a special binomial product and identify its form.

Solution

Since \left(2x + 5\right)\left(2x - 5\right) is a product of a sum and difference, we use the formula and simplify the expression:

\displaystyle \left(a+b\right)\left(a-b\right)\displaystyle =\displaystyle a^{2} - b^{2}Formula for the product of a sum and difference
\displaystyle \left(2x+5\right)\left(2x-5\right)\displaystyle =\displaystyle \left(2x\right)^{2} - 5^{2}Substitute the given values
\displaystyle =\displaystyle 4x^{2} - 25Simplify

Example 4

Multiply and simplify: 3\left(2x + 5y\right)^{2}

Approach

We check first whether 3\left(2x + 5y\right)^{2} involves a special binomial product and identify its form.

Solution

Since \left(2x + 5y\right)^{2} is a square of a binomial in the form \left(a + b\right)^{2}, we use the formula and simplify the expression:

\displaystyle \left(a + b\right)^{2}\displaystyle =\displaystyle a^{2} + 2 a b + b^{2}Formula for the square of a binomial
\displaystyle 3 \left(2x + 5y\right)^{2}\displaystyle =\displaystyle 3 \left[\left(2x\right)^{2} + 2\left(2x\right)\left(5y\right) + \left(5y\right)^{2}\right]Substitute the given values and multiply by 3
\displaystyle =\displaystyle 3 \left[4x^{2} + 20xy + 25y^{2}\right]Distribute the multiplication by 3
\displaystyle =\displaystyle 12x^{2} + 60xy + 75y^{2}Simplify

Outcomes

A1.A.APR.A.1

Add, subtract, and multiply polynomials. Use these operations to demonstrate that polynomials form a closed system that adhere to the same properties of operations as the integers.

A1.MP1

Make sense of problems and persevere in solving them.

A1.MP3

Construct viable arguments and critique the reasoning of others.

A1.MP5

Use appropriate tools strategically.

A1.MP7

Look for and make use of structure.

A1.MP8

Look for and express regularity in repeated reasoning.

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