Topic 1.2 · Unit 1 · CA CCSS A-APR.3, N-CN.9
Factoring and solving higher-degree polynomials
A cubic or quartic will not yield to the quadratic formula, so you find one root, divide it out, and repeat until what is left is quadratic. The rational root theorem tells you which candidates are worth testing.
- Common factor first, always. \( 2x^4 - 18x^2 \) becomes \( 2x^2(x^2 - 9) \) and the rest is easy.
- Recognize the special forms. Sum and difference of cubes: \( a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2) \). Quadratic in disguise: \( x^4 - 5x^2 + 4 \) factors like \( u^2 - 5u + 4 \) with \( u = x^2 \).
- Grouping handles four terms: pair them, factor each pair, and look for the shared bracket.
- Rational root theorem: any rational root is \( \pm\frac{\text{factor of the constant}}{\text{factor of the leading coefficient}} \). Test candidates with the factor theorem, cheapest first.
- Divide out each root found and continue on the quotient. Stop when the quotient is quadratic and finish with factoring or the formula.
- Count the roots. A degree-\( n \) polynomial has exactly \( n \) roots counting multiplicity and complex ones. If you have fewer, you stopped early.
Where marks are lost: solving \( x^4 - 5x^2 + 4 = 0 \) and stopping at \( x^2 = 1, 4 \). Those are values of \( x^2 \), not \( x \); the four solutions are \( \pm 1 \) and \( \pm 2 \).
Example 1: quadratic in disguise. Solve \( x^4 - 13x^2 + 36 = 0 \).
Let \( u = x^2 \): \( u^2 - 13u + 36 = (u - 4)(u - 9) = 0 \), so \( u = 4 \) or \( u = 9 \).
Back-substitute: \( x^2 = 4 \) gives \( x = \pm 2 \); \( x^2 = 9 \) gives \( x = \pm 3 \). Four roots for a quartic, as expected.
Example 2: rational root theorem. Solve \( 2x^3 - 3x^2 - 8x + 12 = 0 \).
Candidates are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \) over \( 1, 2 \). Testing \( x = 2 \): \( 16 - 12 - 16 + 12 = 0 \). ✓
Synthetic division by \( x - 2 \) on \( 2, -3, -8, 12 \) gives \( 2, 1, -6 \), so the quotient is \( 2x^2 + x - 6 = (2x - 3)(x + 2) \).
Roots: \( x = 2, \frac{3}{2}, -2 \).
Example 3: a cubic with complex roots. Solve \( x^3 + 8 = 0 \).
Sum of cubes: \( (x + 2)(x^2 - 2x + 4) = 0 \).
The linear factor gives \( x = -2 \). The quadratic gives \( x = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm 2i\sqrt{3}}{2} = 1 \pm i\sqrt{3} \). Three roots for a cubic, answering only \( x = -2 \) is incomplete.
1–5 factoring, 6–11 solving, 12–15 harder cases and diagnosis.
- Factor \( x^3 - 27 \).
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Difference of cubes with \( a = x \), \( b = 3 \). \( (x-3)(x^2+3x+9) \)
- Factor \( 8x^3 + 125 \).
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\( a = 2x \), \( b = 5 \). \( (2x+5)(4x^2-10x+25) \)
- Factor \( x^4 - 16 \) completely.
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\( (x^2-4)(x^2+4) \), and the first factors again. \( (x-2)(x+2)(x^2+4) \)
- Factor \( x^3 + 2x^2 - 9x - 18 \) by grouping.
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\( x^2(x+2) - 9(x+2) = (x+2)(x^2-9) \). \( (x+2)(x-3)(x+3) \)
- Factor \( 3x^4 - 12x^2 \).
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\( 3x^2(x^2-4) \). \( 3x^2(x-2)(x+2) \)
- Solve \( x^3 - 9x = 0 \).
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\( x(x^2-9) = 0 \). Dividing by \( x \) would lose the root at 0. \( x = 0, 3, -3 \)
- Solve \( x^4 - 5x^2 + 4 = 0 \).
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\( (x^2-1)(x^2-4) = 0 \), so \( x^2 = 1 \) or \( 4 \). \( x = \pm 1, \pm 2 \)
- List the possible rational roots of \( x^3 - 4x^2 + x + 6 \).
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Factors of 6 over factors of 1. \( \pm 1, \pm 2, \pm 3, \pm 6 \)
- Solve \( x^3 - 4x^2 + x + 6 = 0 \).
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\( P(-1) = -1 - 4 - 1 + 6 = 0 \). Dividing by \( x + 1 \) gives \( x^2 - 5x + 6 = (x-2)(x-3) \). \( x = -1, 2, 3 \)
- Solve \( 2x^3 + x^2 - 13x + 6 = 0 \).
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\( P(2) = 16 + 4 - 26 + 6 = 0 \). Dividing gives \( 2x^2 + 5x - 3 = (2x-1)(x+3) \). \( x = 2, \frac{1}{2}, -3 \)
- Solve \( x^3 - 1 = 0 \) completely.
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\( (x-1)(x^2+x+1) = 0 \); the quadratic gives \( x = \frac{-1 \pm i\sqrt{3}}{2} \). \( x = 1, \frac{-1 \pm i\sqrt{3}}{2} \)
- Write a polynomial of least degree with roots \( 2 \), \( -1 \) and \( 3 \).
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\( (x-2)(x+1)(x-3) \); expanding gives \( x^3 - 4x^2 + x + 6 \). \( x^3 - 4x^2 + x + 6 \)
- Write a real-coefficient polynomial of least degree with roots \( 3 \) and \( 2i \).
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Real coefficients force \( -2i \) as well, so \( (x-3)(x^2+4) = x^3 - 3x^2 + 4x - 12 \). \( x^3 - 3x^2 + 4x - 12 \)
- A student solves \( x^4 - 13x^2 + 36 = 0 \) and answers \( x = 4, 9 \). Find the error.
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Those are values of \( u = x^2 \), not of \( x \). Taking square roots of each gives four roots. \( x = \pm 2, \pm 3 \)
- A student solves a cubic and finds only one root. What should that prompt them to check?
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A cubic has exactly three roots counting multiplicity and complex ones, so the quadratic quotient was never solved; it either factors further or needs the formula, possibly giving a conjugate pair. Two roots are missing