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

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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.

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

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

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

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

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

So

$x = - 3, - 7$ $x=-3,-7$

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

How do you solve (5x−25)(x2+10x+21)=0(5x - 25) ( x ^ { 2} + 10x + 21) = 0?