How do you simplify $\frac{c^{2} + c - 56}{c^{2} + 10 c + 16}$ $(c^2+c-56)/(c^2+10c+16)$ ?

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How do you simplify $\frac{c^{2} + c - 56}{c^{2} + 10 c + 16}$ $(c^2+c-56)/(c^2+10c+16)$ ?

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Answer 1 · 2015-10-18T08:21:05

Factor each of the quadratics and eliminate the common factors to find:

$\frac{c^{2} + c - 56}{c^{2} + 10 c + 16} = \frac{c - 7}{c + 2} = 1 - \frac{9}{c + 2}$ $(c^2+c-56)/(c^2+10c+16) = (c-7)/(c+2) = 1 - 9/(c+2)$

with restriction $c \neq - 8$ $c != -8$

Explanation:

$\frac{c^{2} + c - 56}{c^{2} + 10 c + 16} = \frac{(c + 8) (c - 7)}{(c + 8) (c + 2)} = \frac{c - 7}{c + 2}$ $(c^2+c-56)/(c^2+10c+16) = ((c+8)(c-7))/((c+8)(c+2)) = (c-7)/(c+2)$

with restriction $c \neq - 8$ $c != -8$

The restriction is necessary, because if $c = - 8$ $c = -8$ then both the numerator and denominator of the original expression are $0$ $0$ so their quotient is undefined, but the simplified expression $\frac{c - 7}{c + 2}$ $(c-7)/(c+2)$ is defined when $c = - 8$ $c = -8$ .

Also

$\frac{c - 7}{c + 2} = \frac{c + 2 - 2 - 7}{c + 2} = 1 - \frac{9}{c + 2}$ $(c-7)/(c+2) = (c+2-2-7)/(c+2) = 1 - 9/(c+2)$

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How do you simplify $\frac{c^{2} + c - 56}{c^{2} + 10 c + 16}$ $(c^2+c-56)/(c^2+10c+16)$ ?

1 Answer

Explanation:

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How do you simplify c2+c−56c2+10c+16(c^2+c-56)/(c^2+10c+16)?

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How do you simplify $\frac{c^{2} + c - 56}{c^{2} + 10 c + 16}$ $(c^2+c-56)/(c^2+10c+16)$ ?