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

Question

3664 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-27T22:43:56

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

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

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

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

That is:

$x(x^2-49) = 3(x^2-49)$

So either $x = 3$ or $x^2-49 = 0$ , giving $x = +-sqrt(49) = +-7$

$color(white)()$

Alternatively and more systematically:

Subtract the right hand side from the left to get:

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

Let $f(x) = x^3-3x^2-49x+147$

Factor by grouping:

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

$=(x^3-3x^2)-(49x-147)$

$=x^2(x-3)-49(x-3)$

$=(x^2-49)(x-3)$

$=(x^2-7^2)(x-3)$

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

...using the difference of squares identity:

$a^2-b^2 = (a-b)(a+b)$

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

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