How do you solve $(5x - 25) ( x ^ { 2} + 10x + 21) = 0$ ?

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Answer 1 · 2017-04-21T22:33:10

See below.

Since this is already "factored," all we do is set each expression inside the parenthesis equal to zero.

$5x-25=0$
$5x=25$
$x=5$

$x^2+10x+21=0$

$x^2+7x+3x+21=0$

$x(x+7)+3(x+7)=0$

$(x+3)(x+7)=0$

So

$x=-3,-7$

Thus, our solutions for $x$ are $-3,5,-7$ .

How do you solve (5x - 25) ( x ^ { 2} + 10x + 21) = 0(5x - 25) ( x ^ { 2} + 10x + 21) = 0?