H Algebra 2/Trig · Unit 2
Factor a³ + b³ and a³ − b³. On Hard, solve by factoring and use the quadratic formula.
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Sum of cubes
a³ + b³ = (a + b)(a² − ab + b²)
Difference of cubes
a³ − b³ = (a − b)(a² + ab + b²)
Signs
The binomial copies the sign in the problem.
The middle term of the trinomial flips that sign.
The last term is always positive.
Solving
Factor first. The binomial gives one real solution.
The trinomial does not factor. Its discriminant is negative, so use the quadratic formula. Those two solutions are imaginary.
A variable cubed, plus or minus a perfect cube.
The formula is
a³ + b³ = (a + b)(a² − ab + b²)
Substitute a = x and b = 1.
(x + 1)(x² − x·1 + 1²)
(x + 1)(x² − x + 1)
Answer: (x + 1)(x² − x + 1)
The formula is
a³ − b³ = (a − b)(a² + ab + b²)
Substitute a = x and b = 2.
(x − 2)(x² + x·2 + 2²)
(x − 2)(x² + 2x + 4)
Answer: (x − 2)(x² + 2x + 4)
The formula is
a³ + b³ = (a + b)(a² − ab + b²)
Substitute a = y and b = 3.
(y + 3)(y² − y·3 + 3²)
(y + 3)(y² − 3y + 9)
Answer: (y + 3)(y² − 3y + 9)
The formula is
a³ − b³ = (a − b)(a² + ab + b²)
Substitute a = m and b = 4.
(m − 4)(m² + m·4 + 4²)
(m − 4)(m² + 4m + 16)
Answer: (m − 4)(m² + 4m + 16)
The formula is
a³ + b³ = (a + b)(a² − ab + b²)
Substitute a = p and b = 5.
(p + 5)(p² − p·5 + 5²)
(p + 5)(p² − 5p + 25)
Answer: (p + 5)(p² − 5p + 25)
The formula is
a³ − b³ = (a − b)(a² + ab + b²)
Substitute a = n and b = 6.
(n − 6)(n² + n·6 + 6²)
(n − 6)(n² + 6n + 36)
Answer: (n − 6)(n² + 6n + 36)
A coefficient, two variables, or a greatest common factor.
The formula is
a³ + b³ = (a + b)(a² − ab + b²)
Substitute a = 2x and b = 3.
(2x + 3)((2x)² − (2x)·3 + 3²)
(2x + 3)(4x² − 6x + 9)
Answer: (2x + 3)(4x² − 6x + 9)
The formula is
a³ − b³ = (a − b)(a² + ab + b²)
Substitute a = 3x and b = 1.
(3x − 1)((3x)² + (3x)·1 + 1²)
(3x − 1)(9x² + 3x + 1)
Answer: (3x − 1)(9x² + 3x + 1)
The formula is
a³ + b³ = (a + b)(a² − ab + b²)
Substitute a = 4p and b = 5.
(4p + 5)((4p)² − (4p)·5 + 5²)
(4p + 5)(16p² − 20p + 25)
Answer: (4p + 5)(16p² − 20p + 25)
The formula is
a³ − b³ = (a − b)(a² + ab + b²)
Substitute a = x and b = 2y.
(x − 2y)(x² + x·(2y) + (2y)²)
(x − 2y)(x² + 2xy + 4y²)
Answer: (x − 2y)(x² + 2xy + 4y²)
The formula is
a³ + b³ = (a + b)(a² − ab + b²)
Substitute a = 2m and b = n.
(2m + n)((2m)² − (2m)·n + n²)
(2m + n)(4m² − 2mn + n²)
Answer: (2m + n)(4m² − 2mn + n²)
The GCF is 5.
5(t³ − 8)
The formula is
a³ − b³ = (a − b)(a² + ab + b²)
Substitute a = t and b = 2.
5(t − 2)(t² + t·2 + 2²)
5(t − 2)(t² + 2t + 4)
Answer: 5(t − 2)(t² + 2t + 4)
Factor, then use the quadratic formula. The trinomial does not factor.
The formula is
a³ + b³ = (a + b)(a² − ab + b²)
Substitute a = x and b = 2.
(x + 2)(x² − x·2 + 2²)
(x + 2)(x² − 2x + 4)
Set each factor equal to zero.
x + 2 = 0
x = −2
The trinomial does not factor. Use the quadratic formula.
x² − 2x + 4 = 0
x = −b ± √(b² − 4ac)2a
x = −(−2) ± √((−2)² − 4(1)(4))2
x = 2 ± √(4 − 16)2
x = 2 ± √(−12)2
x = 2 ± i√122
x = 2 ± 2i√32
x = 1 ± i√3
Answer: x = −2 and x = 1 ± i√3
The formula is
a³ − b³ = (a − b)(a² + ab + b²)
Substitute a = x and b = 3.
(x − 3)(x² + x·3 + 3²)
(x − 3)(x² + 3x + 9)
Set each factor equal to zero.
x − 3 = 0
x = 3
The trinomial does not factor. Use the quadratic formula.
x² + 3x + 9 = 0
x = −b ± √(b² − 4ac)2a
x = −3 ± √(3² − 4(1)(9))2
x = −3 ± √(9 − 36)2
x = −3 ± √(−27)2
x = −3 ± i√272
x = −3 ± 3i√32
Answer: x = 3 and x = −3 ± 3i√32
The formula is
a³ + b³ = (a + b)(a² − ab + b²)
Substitute a = x and b = 4.
(x + 4)(x² − x·4 + 4²)
(x + 4)(x² − 4x + 16)
Set each factor equal to zero.
x + 4 = 0
x = −4
The trinomial does not factor. Use the quadratic formula.
x² − 4x + 16 = 0
x = −b ± √(b² − 4ac)2a
x = −(−4) ± √((−4)² − 4(1)(16))2
x = 4 ± √(16 − 64)2
x = 4 ± √(−48)2
x = 4 ± i√482
x = 4 ± 4i√32
x = 2 ± 2i√3
Answer: x = −4 and x = 2 ± 2i√3
The formula is
a³ − b³ = (a − b)(a² + ab + b²)
Substitute a = 2x and b = 1.
(2x − 1)((2x)² + (2x)·1 + 1²)
(2x − 1)(4x² + 2x + 1)
Set each factor equal to zero.
2x − 1 = 0
2x = 1
x = 12
The trinomial does not factor. Use the quadratic formula.
4x² + 2x + 1 = 0
x = −b ± √(b² − 4ac)2a
x = −2 ± √(2² − 4(4)(1))8
x = −2 ± √(4 − 16)8
x = −2 ± √(−12)8
x = −2 ± i√128
x = −2 ± 2i√38
x = −1 ± i√34
Answer: x = 12 and x = −1 ± i√34
The formula is
a³ + b³ = (a + b)(a² − ab + b²)
Substitute a = 3x and b = 1.
(3x + 1)((3x)² − (3x)·1 + 1²)
(3x + 1)(9x² − 3x + 1)
Set each factor equal to zero.
3x + 1 = 0
3x = −1
x = −13
The trinomial does not factor. Use the quadratic formula.
9x² − 3x + 1 = 0
x = −b ± √(b² − 4ac)2a
x = −(−3) ± √((−3)² − 4(9)(1))18
x = 3 ± √(9 − 36)18
x = 3 ± √(−27)18
x = 3 ± i√2718
x = 3 ± 3i√318
x = 1 ± i√36
Answer: x = −13 and x = 1 ± i√36
The formula is
a³ + b³ = (a + b)(a² − ab + b²)
Substitute a = 5x and b = 2.
(5x + 2)((5x)² − (5x)·2 + 2²)
(5x + 2)(25x² − 10x + 4)
Set each factor equal to zero.
5x + 2 = 0
5x = −2
x = −25
The trinomial does not factor. Use the quadratic formula.
25x² − 10x + 4 = 0
x = −b ± √(b² − 4ac)2a
x = −(−10) ± √((−10)² − 4(25)(4))50
x = 10 ± √(100 − 400)50
x = 10 ± √(−300)50
x = 10 ± i√30050
x = 10 ± 10i√350
x = 1 ± i√35
Answer: x = −25 and x = 1 ± i√35
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