# How do you solve x^3 + 147 = 3x^2 + 49x?

Nov 27, 2015

Subtract $147 + 49 x$ from both sides, then identify common factors on each side to find:

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

#### Explanation:

Subtract $147 + 49 x$ from both sides to get:

${x}^{3} - 49 x = 3 {x}^{2} - 147$

That is:

$x \left({x}^{2} - 49\right) = 3 \left({x}^{2} - 49\right)$

So either $x = 3$ or ${x}^{2} - 49 = 0$, giving $x = \pm \sqrt{49} = \pm 7$

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Alternatively and more systematically:

Subtract the right hand side from the left to get:

${x}^{3} - 3 {x}^{2} - 49 x + 147 = 0$

Let $f \left(x\right) = {x}^{3} - 3 {x}^{2} - 49 x + 147$

Factor by grouping:

${x}^{3} - 3 {x}^{2} - 49 x + 147$

$= \left({x}^{3} - 3 {x}^{2}\right) - \left(49 x - 147\right)$

$= {x}^{2} \left(x - 3\right) - 49 \left(x - 3\right)$

$= \left({x}^{2} - 49\right) \left(x - 3\right)$

$= \left({x}^{2} - {7}^{2}\right) \left(x - 3\right)$

$= \left(x - 7\right) \left(x + 7\right) \left(x - 3\right)$

...using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

So $f \left(x\right) = 0$ has roots $x = 7$, $x = - 7$ and $x = 3$