How do you factor $y = 7 x^{3} + 28 x^{2}$ $y= 7x^3+28x^2$ ?

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Answer 1 · 2015-12-25T11:01:52

Both $7 x^{3}$ $7x^3$ and $28 x^{2}$ $28x^2$ are divisible by $7 x^{2}$ $7x^2$ , so:

$7 x^{3} + 28 x^{2} = 7 x^{2} (x + 4)$ $7x^3+28x^2 = 7x^2(x+4)$

$7 x^{3} = 7 x^{2} \cdot x$ $7x^3 = 7x^2*x$

$28 x^{2} = 7 x^{2} \cdot 4$ $28x^2 = 7x^2*4$

So:

$7 x^{3} + 28 x^{2} = (7 x^{2} \cdot x) + (7 x^{2} \cdot 4) = 7 x^{2} \cdot (x + 4)$ $7x^3+28x^2 = (7x^2*x)+(7x^2*4) = 7x^2*(x+4)$

How do you factor y= 7x^3+28x^2y=7x3+28x2y= 7x^3+28x^2 ?