How do you solve $(x - 1) (x + 4) = 14$ $(x-1)(x+4)=14$ ?

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How do you solve $(x - 1) (x + 4) = 14$ $(x-1)(x+4)=14$ ?

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Answer 1 · 2017-02-08T22:53:36

The answer is $x = - 6 and 3$ $x=-6 and 3$

Explanation:

First, use the double distributive property (FOIL method) to multiply $(x - 1)$ $(x-1)$ and $(x + 4)$ $(x+4)$ together, which gives you $x^{2}$ $x^2$ $+$ $+$ $4 x$ $4x$ $-$ $-$ $x$ $x$ $-$ $-$ $4$ $4$ $=$ $=$ $14$ $14$

Simplify the like terms: $x^{2}$ $x^2$ $+$ $+$ $3 x$ $3x$ $-$ $-$ $4$ $4$ $=$ $=$ $14$ $14$

Next, set the equation equal to zero by subtracting 14 from both sides: $x^{2}$ $x^2$ $+$ $+$ $3 x$ $3x$ $-$ $-$ $18$ $18$ $=$ $=$ $0$ $0$

Then, factor the equation: $(x + 6) (x - 3)$ $(x+6)(x-3)$ $=$ $=$ $0$ $0$

Set each quantity each to zero and solve for $x$ $x$

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How do you solve $(x - 1) (x + 4) = 14$ $(x-1)(x+4)=14$ ?

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How do you solve $(x - 1) (x + 4) = 14$ $(x-1)(x+4)=14$ ?