$$(4x^{3}-7x^{2}-45x+36)\div(x+3)$$
$\left(x-4\right)\left(4x-3\right)$
$$\frac{4\left(\frac{1}{4}x-1\right)\left(4x-3\right)\left(x+3\right)}{x+3}$$
$$4\left(\frac{1}{4}x-1\right)\left(4x-3\right)$$
$$4x^{2}-19x+12$$
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$4x^{2}-19x+12$