Multiply a Monomial and a Polynomial
To multiply a monomial and a polynomial with two or more terms, apply the distributive property.
Distributive Property: a(b + c) = ab + ac
Example 1. Find each product.
a) 2x²(4x – 2)
b) 8ab³(2a – 3b + c)
Solution.
a) 2x²(4x – 2) = 8x 3– 4x² b)
8ab³(2a – 3b + c) = 16a 2b³ – 24ab4 + 8ab³c
Multiply Binomials
To multiply binomials, apply the distributive property twice. First, multiply the binomial and each term of the polynomial. Then combine like terms, if possible.
(x – 3)(x 2 – 2x + 1)
= (x – 3)x 2 + (x – 3)(– 2x) + (x – 3)(1)
= x 3 – 3x 2 – 2x 2 + 6x + x – 3
= x 3 – 5x 2 + 7x – 3
Example 2. Find each product.
a) (6x – 5)(4x + 2)
b) (2x – 1)(4x 2 + 2x + 1)
c) (x + y)(z – 3)
Solution. a)
(6x – 5)(4x + 2)
= (6x – 5)(4x) + (6x – 5)(2)
= 24x 2 – 20x + 12x – 10
= 24x 2 – 8x – 10
b)
(2x – 1)(4x 2 + 2x + 1)
= (2x – 1)(4x 2) + (2x – 1)(2x) + (2x – 1)(1)
= 8x 3 – 4x 2 + 4x 2 – 2x + 2x – 1
= 8x 3 – 1
c)
(x + y)(z – 3)
= (x + y)(z) + (x + y)(–3)
= xz + yz – 3x – 3y
Helpful Formulas
Formulas can be helpful when multiplying polynomials. In this section we will consider five such formulas.
Square of a Sum (a + b)2 = a2 + 2ab + b2 Square of a Difference
(a – b)2 = a 2 – 2ab + b2
Difference of Squares
(a + b)(a – b) = a 2 – b2
Difference of Cubes (a – b)(a2 + ab + b2 ) = a3 – b3 Sum of Cubes (a + b)(a2 – ab + b2 ) = a 3 + b 3
Each of these formulas can be easily verified by multiplying them out term–by–term. The formula for Difference of Cubes is shown below. You are urged to try the others yourself.
Verify the formula for the Difference of Cubes.
(a – b)(a2 + 2ab + b2)
= (a – b)(a2) + (a – b)(ab) + (a – b)(b2)
= a3– a2b + a2b – ab2 + ab2– b3
= a 3– b 3
Find each product.
a) (2x – 1)2
b) (3x – 4)(3x + 4)
c) (2x + 3y)3
d) (3x – 2)(9x2 + 6x + 4)
Solution.
a) (2x – 1)2
= (2x)2 – 2(2x)(1) + (1)2
= 4x2 – 4x + 1
Use the formula for the square of a difference.
(a – b) 2 = a 2 – 2ab + b 2
b) (3x – 4)(3x + 4)
= (3x)2 – (4)2
=9x2 – 16
This is a difference of squares.
(a + b)(a – b) = a2 – b2
c) (2x + 3y)3
= (2x + 3y)(2x + 3y)2
= (2x + 3y)[(2x)2 + 2(2x)(3y) + (3y)2 ]
= (2x + 3y)[4x2 + 12xy + 9y2 ]
= (2x + 3y)[4x2 ] + (2x + 3y)[12xy] + (2x + 3y)[9y 2]
= 8x3 + 12x2 y + 24x2 y + 36xy2 + 18xy2 + 27y 3
= 8x3 + 36x2 y + 54xy2 + 27y3
This is a cube of a sum. We can break the product up so that we can apply the formula for the square of a sum.
d) (3x – 2)(9x2 + 6x + 4)
= (3x – 2)[(3x)2 + (2)(3)x + 22)
= (3x)3– 23
= 27x3– 8
This is a difference of cubes.
(a – b)(a2 + ab + b2 ) = a3– b3
0 out of 0 correct.