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Solve quadratic equations by factoring, including shifted forms

Move everything to one side

To solve a quadratic by factoring, first rewrite it as $a x^{2} + b x + c = 0$ . Remove any common numerical factor, then write the polynomial as a product. Each factor can be zero separately.

Problem

Solve by factoring: (1) $x^{2} + 5 x - 24 = 0$ ; (2) $n^{2} + 12 n + 32 = 0$ ; (3) $140 n^{2} + 77 n - 32 = 5 n$ ; (4) $47 x^{2} + 236 x - 33 = - 2 x - 2 x^{2} + 2$ .

Answer

(1)

x = - 8, 3

; (2)

n = - 8, - 4

; (3)

n = 2/7, - 4/5

; (4)

x = 1/7, - 5

The equations with terms on both sides must be rearranged before factoring.

Step-by-step solution

1. Factor the first two

$x^{2} + 5 x - 24 = (x + 8) (x - 3)$ , so $x = - 8$ or $3$ . Also $n^{2} + 12 n + 32 = (n + 8) (n + 4)$ , so $n = - 8$ or $- 4$ .

2. Rearrange the third

Moving $5 n$ left and dividing by 4 gives $35 n^{2} + 18 n - 8 = 0$ . This factors as $(7 n - 2) (5 n + 4) = 0$ , giving $n = 2/7$ or $- 4/5$ .

3. Rearrange the fourth

Moving every term left and dividing by 7 gives $7 x^{2} + 34 x - 5 = 0$ . Its factorization is $(7 x - 1) (x + 5) = 0$ , giving $x = 1/7$ or $- 5$ .

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