EXAMPLE 1
Find a common monomial factor
Factor the polynomial completely.
a.
b.
x3 + 2x2 – 15x = x(x2 + 2x – 15)
Factor common
monomial.
= x(x + 5)(x – 3)
Factor trinomial.
2y5 – 18y3 = 2y3(y2 – 9)
= 2y3(y + 3)(y – 3)
c.
Factor common
monomial.
Difference of two
squares
4z4 – 16z3 + 16z2 = 4z2(z2 – 4z + 4) Factor common
monomial.
= 4z2(z – 2)2
Perfect square trinomial
EXAMPLE 2
Factor the sum or difference of two cubes
Factor the polynomial completely.
a.
x3 + 64= x3 + 43
Sum of two cubes
= (x + 4)(x2 – 4x + 16)
b.
16z5 – 250z2 = 2z2(8z3 – 125)
= 2z2 (2z)3 – 53
Factor common monomial.
Difference of two cubes
= 2z2(2z – 5)(4z2 + 10z + 25)
EXAMPLE 3
Factor by grouping
Factor the polynomial x3 – 3x2 – 16x + 48 completely.
x3 – 3x2 – 16x + 48 = x2(x – 3) – 16(x – 3) Factor by grouping.
= (x2 – 16)(x – 3)
Distributive property
= (x + 4)(x – 4)(x – 3) Difference of two
squares
EXAMPLE 4
Factor polynomials in quadratic form
Factor completely: (a) 16x4 – 81 and (b) 2p8 + 10p5 + 12p2.
a. 16x4 – 81 = (4x2)2 – 92
Write as difference
of two squares.
= (4x2 + 9)(4x2 – 9)
Difference of two
squares
= (4x2 + 9)(2x + 3)(2x – 3)
Difference of two
squares
b. 2p8 + 10p5 + 12p2 = 2p2(p6 + 5p3 + 6)
= 2p2(p3 + 3)(p3 + 2)
Factor common
monomial.
Factor trinomial in
quadratic form.
EXAMPLE 5
Standardized Test Practice
SOLUTION
3x5 + 15x = 18x3
3x5 – 18x3 + 15x = 0
3x(x4 – 6x2 + 5) = 0
Write original equation.
Write in standard form.
Factor common monomial.
EXAMPLE 5
Standardized Test Practice
3x(x2 – 1)(x2 – 5) = 0
Factor trinomial.
3x(x + 1)(x – 1)(x2 – 5) = 0
Difference of two
squares
x = 0, x = – 1, x = 1, x =
5 ,or x = –
5
ANSWER The correct answer is D.
Zero product
property
EXAMPLE 1
Use polynomial long division
Divide f (x) = 3x4 – 5x3 + 4x – 6 by x2 – 3x + 5.
SOLUTION
Write polynomial division in the same format you
use when dividing numbers. Include a “0” as the
coefficient of x2 in the dividend. At each stage,
divide the term with the highest power in what is
left of the dividend by the first term of the divisor.
This gives the next term of the quotient.
EXAMPLE 1
Use polynomial long division
3x2 + 4x – 3
x2 – 3x + 5 )3x4 – 5x3 + 0x2 + 4x – 6
quotient
Multiply divisor by 3x4/x2 = 3x2
3x4 – 9x3 + 15x2
4x3 – 15x2 + 4x
Subtract.
Bring down next term.
4x3 – 12x2 + 20x
Multiply divisor by 4x3/x2 = 4x
– 3x2 – 16x – 6
–3x2 + 9x – 15
– 25x + 9
Subtract.
Bring down next term.
Multiply divisor by – 3x2/x2 = – 3
remainder
EXAMPLE 1
ANSWER
Use polynomial long division
3x4 – 5x3 + 4x – 6 = 3x2 + 4x – 3 + – 25x + 9
x2 – 3x + 5
x2 – 3x + 5
CHECK
You can check the result of a division problem by
multiplying the quotient by the divisor and adding the
remainder. The result should be the dividend.
(3x2 + 4x – 3)(x2 – 3x + 5) + (– 25x + 9)
= 3x2(x2 – 3x + 5) + 4x(x2 – 3x + 5) – 3(x2 – 3x + 5) – 25x + 9
= 3x4 – 9x3 + 15x2 + 4x3 – 12x2 + 20x – 3x2 + 9x – 15 – 25x + 9
= 3x4 – 5x3 + 4x – 6
EXAMPLE 2
Use polynomial long division with a linear divisor
Divide f (x) = x3 + 5x2 – 7x + 2 by x – 2.
x2 + 7x + 7
x – 2 ) x3 + 5x2 – 7x + 2
x3 – 2x2
7x2 – 7x
7x2 – 14x
7x + 2
7x – 14
16
ANSWER
quotient
Multiply divisor by x3/x = x2.
Subtract.
Multiply divisor by 7x2/x = 7x.
Subtract.
Multiply divisor by 7x/x = 7.
remainder
x3 + 5x2 – 7x +2
= x2 + 7x + 7 + 16
x–2
x–2
EXAMPLE 3
Use synthetic division
Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic
division.
SOLUTION
–3
ANSWER
1
–8
5
–6
15
– 21
2 –5
7
– 16
2
2x3 + x2 – 8x + 5
16
= 2x2 – 5x + 7 –
x+3
x+3
EXAMPLE 4
Factor a polynomial
Factor f (x) = 3x3 – 4x2 – 28x – 16 completely given that
x + 2 is a factor.
SOLUTION
Because x + 2 is a factor of f (x), you know that
f (– 2) = 0. Use synthetic division to find the other
factors.
– 2 3 – 4 – 28 – 16
–6
20
16
3 – 10
–8
0
EXAMPLE 4
Factor a polynomial
Use the result to write f (x) as a product of two
factors and then factor completely.
f (x) = 3x3 – 4x2 – 28x – 16
Write original polynomial.
= (x + 2)(3x2 – 10x – 8)
Write as a product of two
factors.
= (x + 2)(3x + 2)(x – 4)
Factor trinomial.
EXAMPLE 5
Standardized Test Practice
SOLUTION
Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic
division.
3
1
1
–2
– 23
60
3
3
– 60
1
– 20
0
EXAMPLE 5
Standardized Test Practice
Use the result to write f (x) as a product of two
factors. Then factor completely.
f (x) = x3 – 2x2 – 23x + 60
= (x – 3)(x2 + x – 20)
= (x – 3)(x + 5)(x – 4)
The zeros are 3, – 5, and 4.
ANSWER The correct answer is A.