Factoring Polynomials

Advanced factoring techniques

Factor by Grouping

Group terms and factor out common factors.

Example: $x^3 + 2x^2 + 3x + 6$ $= x^2(x + 2) + 3(x + 2)$ $= (x + 2)(x^2 + 3)$

Sum of cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Difference of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Example: $x^3 - 8$ $= x^3 - 2^3 = (x - 2)(x^2 + 2x + 4)$

Sometimes substituting can simplify factoring.

Example: $x^4 - 5x^2 + 4$

Let $u = x^2$ : $u^2 - 5u + 4 = (u - 4)(u - 1)$

Substitute back: $= (x^2 - 4)(x^2 - 1)$ $= (x + 2)(x - 2)(x + 1)(x - 1)$

Factor by grouping: $x^3 + 3x^2 + 2x + 6$

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Group the first two and last two terms: $(x^3 + 3x^2) + (2x + 6)$

Factor out common factors: $x^2(x + 3) + 2(x + 3)$

Factor out $(x + 3)$ : $(x + 3)(x^2 + 2)$

Answer: $(x + 3)(x^2 + 2)$

Factor: $x^3 + 27$

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This is a sum of cubes: $x^3 + 3^3$

Use the formula: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Here $a = x$ and $b = 3$ : $x^3 + 27 = (x + 3)(x^2 - 3x + 9)$

Answer: $(x + 3)(x^2 - 3x + 9)$

Factor completely: $x^4 - 13x^2 + 36$

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This is a quadratic in form. Let $u = x^2$ : $u^2 - 13u + 36$

Factor: $(u - 9)(u - 4)$

Substitute back $u = x^2$ : $(x^2 - 9)(x^2 - 4)$

Both are difference of squares: $= (x + 3)(x - 3)(x + 2)(x - 2)$

Answer: $(x + 3)(x - 3)(x + 2)(x - 2)$

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