Factoring runs opposite to expanding: it repackages a polynomial as a product.
Just as you recover a rectangle's sides from its area, reading x² + (a+b)x + ab means finding two numbers with sum (a+b) and product ab, giving (x+a)(x+b).
Even x⁴ becomes a familiar quadratic once you substitute t = x², and the factor theorem cracks bigger ones by finding a root.
Slide a, b, and k to explore the area model and the substitution step.
Factoring = Reverse of Multiplication
Factoring is the reverse of expansion. Like finding the side lengths of a rectangle from its area.
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Area model: x² + (a+b)x + ab = (x+a)(x+b)
💡 Find Sum & Product
①In x² + px + q, find two numbers with sum=p, product=q
②Those numbers are a, b → (x+a)(x+b)
③Area model: find the two side lengths
Factoring Formulas
Formulas
Core Factoring Formulas
Factoring by Substitution
Complex expressions become quadratic with the right substitution. For x⁴-form, substitute t = x²; for any repeated block, use one new variable.
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x⁴ + kx² + (k-2) with substitution t = x²
🔑 Strategy
①Spot common parts → substitute one variable
②x⁴ + ax² + b → t = x² → quadratic in t
③Factor, then revert to the original variable
Practical Strategy
Common Factors First
Common factor → apply formula → try substitution
3-step approach
Difference of Squares
a² - b² = (a + b)(a - b)
The most-used factoring formula
Work It Out
Example 1
Factor x³ + 8.
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Since 8 = 2³, use the sum-of-cubes formula a³ + b³ = (a + b)(a² − ab + b²).
a³ + b³ = (a + b)(a² − ab + b²)
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Substitute a = x, b = 2.
x³ + 2³ = (x + 2)(x² − 2x + 4)
▸ (x + 2)(x² − 2x + 4)
Knowing the sum and difference of cubes lets you factor at once.
Example 2
Factor x⁴ − 13x² + 36.
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Let x² = t to get a quadratic.
t² − 13t + 36 = (t − 4)(t − 9)
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Restore t = x² and factor each by difference of squares.
(x² − 4)(x² − 9) = (x − 2)(x + 2)(x − 3)(x + 3)
▸ (x − 2)(x + 2)(x − 3)(x + 3)
For a biquadratic, substitute, then finish with difference of squares.
Wrap-up
Quadratic Factoring
x² + (a+b)x + ab = (x + a)(x + b)
Find two factors via sum & product
Cube Factoring
a³ ± b³ = (a ± b)(a² ∓ ab + b²)
Note: middle term has opposite sign
Grade-10 school exam type
Factor the polynomial x³ − 2x² − 5x + 6.
①(x − 1)(x − 2)(x + 3)
②(x − 1)(x + 2)(x − 3)
③(x + 1)(x − 2)(x − 3)
④(x − 1)(x − 2)(x − 3)
⑤(x + 1)(x + 2)(x − 3)
▸ ② (x − 1)(x + 2)(x − 3)
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Since substituting x = 1 gives 0, by the factor theorem x − 1 is a factor.
f(1) = 1 − 2 − 5 + 6 = 0
2
Divide to get the quotient, then factor again.
(x − 1)(x² − x − 6) = (x − 1)(x + 2)(x − 3)
🎯 Exam Points
①Pull out common factors first
②Difference of squares: check if both terms are perfect squares
③Perfect square trinomial: middle term must be 2ab
④Substitution: try t = x² for x⁴ problems
⑤Use Factor Theorem: try divisors of the constant term to find roots