H Algebra 2/Trig · Unit 3
Factor completely, then solve each factor. List real and imaginary zeros.
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Common factor first
Factor out the greatest common factor, including any power of x. A number alone is never zero.
A fourth power
After you factor out x, let u = x². Solve for u, then take square roots.
Factor, then solve x² = a number.
Factor out the common factor.
6x³(x² + 4) = 0
Set each factor that can be zero equal to 0.
x³ = 0
x = 0, multiplicity 3
x² + 4 = 0
x² = −4
A negative under the square root gives an imaginary number.
x = 2i or x = −2i
Answer: x = 0, multiplicity 3
x = 2i or x = −2i
Factor out the common factor.
4x²(x² + 3) = 0
Set each factor that can be zero equal to 0.
x² = 0
x = 0, multiplicity 2
x² + 3 = 0
x² = −3
A negative under the square root gives an imaginary number.
x = i or x = −i
Answer: x = 0, multiplicity 2
x = i or x = −i
Factor out the common factor.
3x³(x² − 16) = 0
Set each factor that can be zero equal to 0.
x³ = 0
x = 0, multiplicity 3
x² − 16 = 0
x² = 16
x = 4 or x = −4
Answer: x = 0, multiplicity 3
x = 4 or x = −4
Factor out the common factor.
5x(x² + 4) = 0
Set each factor that can be zero equal to 0.
x = 0
x = 0
x² + 4 = 0
x² = −4
A negative under the square root gives an imaginary number.
x = 2i or x = −2i
Answer: x = 0
x = 2i or x = −2i
Factor out the common factor.
2x²(x² − 9) = 0
Set each factor that can be zero equal to 0.
x² = 0
x = 0, multiplicity 2
x² − 9 = 0
x² = 9
x = 3 or x = −3
Answer: x = 0, multiplicity 2
x = 3 or x = −3
Factor out the common factor.
8x³(x² + 1) = 0
Set each factor that can be zero equal to 0.
x³ = 0
x = 0, multiplicity 3
x² + 1 = 0
x² = −1
A negative under the square root gives an imaginary number.
x = i or x = −i
Answer: x = 0, multiplicity 3
x = i or x = −i
Factor x, then factor the degree-4 piece.
Factor out the common factor of x.
x(x⁴ + 2x² − 15) = 0
x = 0, or the degree-4 factor is 0.
x = 0
Let u = x².
u⁴ + 2u² − 15 = 0
(u + 5)(u − 3) = 0
u = −5 or u = 3
Put x² back in for u.
x² = −5
A negative under the square root gives an imaginary number.
x = i or x = −i
x² = 3
x = or x = −
Answer: x = 0
x = i or x = −i
x = or x = −
Factor out the common factor of x.
x(x⁴ − 10x² + 9) = 0
x = 0, or the degree-4 factor is 0.
x = 0
Let u = x².
u⁴ − 10u² + 9 = 0
(u − 9)(u − 1) = 0
u = 9 or u = 1
Put x² back in for u.
x² = 9
x = 3 or x = −3
x² = 1
x = 1 or x = −1
Answer: x = 0
x = 3 or x = −3
x = 1 or x = −1
Factor out the common factor of x.
x(x⁴ + 5x² − 36) = 0
x = 0, or the degree-4 factor is 0.
x = 0
Let u = x².
u⁴ + 5u² − 36 = 0
(u + 9)(u − 4) = 0
u = −9 or u = 4
Put x² back in for u.
x² = −9
A negative under the square root gives an imaginary number.
x = 3i or x = −3i
x² = 4
x = 2 or x = −2
Answer: x = 0
x = 3i or x = −3i
x = 2 or x = −2
Factor out the common factor of x.
2x(x⁴ − 4x² − 12) = 0
The constant 2 is never 0.
x = 0
Let u = x².
u⁴ − 4u² − 12 = 0
(u − 6)(u + 2) = 0
u = 6 or u = −2
Put x² back in for u.
x² = 6
x = or x = −
x² = −2
A negative under the square root gives an imaginary number.
x = i or x = −i
Answer: x = 0
x = or x = −
x = i or x = −i
Factor out the common factor of x.
x(x⁴ − 13x² + 36) = 0
x = 0, or the degree-4 factor is 0.
x = 0
Let u = x².
u⁴ − 13u² + 36 = 0
(u − 9)(u − 4) = 0
u = 9 or u = 4
Put x² back in for u.
x² = 9
x = 3 or x = −3
x² = 4
x = 2 or x = −2
Answer: x = 0
x = 3 or x = −3
x = 2 or x = −2
Factor out the common factor of x.
x(x⁴ + x² − 12) = 0
x = 0, or the degree-4 factor is 0.
x = 0
Let u = x².
u⁴ + u² − 12 = 0
(u + 4)(u − 3) = 0
u = −4 or u = 3
Put x² back in for u.
x² = −4
A negative under the square root gives an imaginary number.
x = 2i or x = −2i
x² = 3
x = or x = −
Answer: x = 0
x = 2i or x = −2i
x = or x = −
The quadratic in u has a leading coefficient greater than 1.
Factor out x.
x(2x⁴ − 5x² − 12) = 0
x = 0
Let u = x².
2u² − 5u − 12 = 0
Find two numbers that multiply to 2 · −12 = −24 and add to −5: −8 and 3.
2u² − 8u + 3u − 12
Group.
2u(u − 4) + 3(u − 4) = 0
(2u + 3)(u − 4) = 0
2u + 3 = 0
u = −32
u − 4 = 0
u = 4
Put x² back in for u.
x² = −32
A negative under the square root gives an imaginary number.
Rationalize the denominator.
x = i2 or x = −i2
x² = 4
x = 2 or x = −2
Answer: x = 0
x = i2 or x = −i2
x = 2 or x = −2
Factor out x.
x(2x⁴ + x² − 15) = 0
x = 0
Let u = x².
2u² + u − 15 = 0
Find two numbers that multiply to 2 · −15 = −30 and add to 1: 6 and −5.
2u² + 6u − 5u − 15
Group.
2u(u + 3) − 5(u + 3) = 0
(2u − 5)(u + 3) = 0
2u − 5 = 0
u = 52
u + 3 = 0
u = −3
Put x² back in for u.
x² = 52
Rationalize the denominator.
x = 2 or x = −2
x² = −3
A negative under the square root gives an imaginary number.
x = i or x = −i
Answer: x = 0
x = 2 or x = −2
x = i or x = −i
Factor out x.
x(2x⁴ − 11x² + 12) = 0
x = 0
Let u = x².
2u² − 11u + 12 = 0
Find two numbers that multiply to 2 · 12 = 24 and add to −11: −8 and −3.
2u² − 8u − 3u + 12
Group.
2u(u − 4) − 3(u − 4) = 0
(2u − 3)(u − 4) = 0
2u − 3 = 0
u = 32
u − 4 = 0
u = 4
Put x² back in for u.
x² = 32
Rationalize the denominator.
x = 2 or x = −2
x² = 4
x = 2 or x = −2
Answer: x = 0
x = 2 or x = −2
x = 2 or x = −2
Factor out x.
x(3x⁴ − 16x² − 12) = 0
x = 0
Let u = x².
3u² − 16u − 12 = 0
Find two numbers that multiply to 3 · −12 = −36 and add to −16: −18 and 2.
3u² − 18u + 2u − 12
Group.
3u(u − 6) + 2(u − 6) = 0
(3u + 2)(u − 6) = 0
3u + 2 = 0
u = −23
u − 6 = 0
u = 6
Put x² back in for u.
x² = −23
A negative under the square root gives an imaginary number.
Rationalize the denominator.
x = i3 or x = −i3
x² = 6
x = or x = −
Answer: x = 0
x = i3 or x = −i3
x = or x = −
Factor out x.
x(2x⁴ − 15x² − 8) = 0
x = 0
Let u = x².
2u² − 15u − 8 = 0
Find two numbers that multiply to 2 · −8 = −16 and add to −15: −16 and 1.
2u² − 16u + u − 8
Group.
2u(u − 8) + (u − 8) = 0
(2u + 1)(u − 8) = 0
2u + 1 = 0
u = −12
u − 8 = 0
u = 8
Put x² back in for u.
x² = −12
A negative under the square root gives an imaginary number.
Rationalize the denominator.
x = i2 or x = −i2
x² = 8
x = 2 or x = −2
Answer: x = 0
x = i2 or x = −i2
x = 2 or x = −2
Factor out x.
x(3x⁴ + 2x² − 8) = 0
x = 0
Let u = x².
3u² + 2u − 8 = 0
Find two numbers that multiply to 3 · −8 = −24 and add to 2: 6 and −4.
3u² + 6u − 4u − 8
Group.
3u(u + 2) − 4(u + 2) = 0
(3u − 4)(u + 2) = 0
3u − 4 = 0
u = 43
u + 2 = 0
u = −2
Put x² back in for u.
x² = 43
Rationalize the denominator.
x = 23 or x = −23
x² = −2
A negative under the square root gives an imaginary number.
x = i or x = −i
Answer: x = 0
x = 23 or x = −23
x = i or x = −i
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