The product $(x + 2) (x^{2} - 2 x + 4)$ is equal to

x^{3} + 8

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x^{3} - 8

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x^{3} - 4 x^{2} + 4 x - 8

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x^{3} + 4 x^{2} + 2 x + 8

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Solution

The correct option is A $x^{3} + 8$
$(x + 2) (x^{2} - 2 x + 4)$

= $x (x^{2} - 2 x + 4) + 2 (x^{2} - 2 x + 4)$

= $x^{3} - 2 x^{2} + 4 x + 2 x^{2} - 4 x + 8$

= $x^{3} + 8$

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Similar questions

x^{2} + 2 x - 8, x^{3} - 4 x^{2} + 4 x

x^{2} - 4

What is the H.C.F of $x^{2} + 2 x - 8, x^{3} - 4 x^{2} + 4 x, x^{2} - 4$

i .

lim x \to 1 [\frac{x^{2} + 1}{x + 100}]

i i .

lim x \to 2 [\frac{x^{3} - 4 x^{2} + 4 x}{x^{2} - 4}]

i i i .

lim x \to 2 [\frac{x^{2} - 4}{x^{3} - 4 x^{2} + 4 x}]

i v .

lim x \to 2 [\frac{x^{3} - 2 x^{2}}{x^{2} - 5 x + 6}]

v .

lim x \to 1 [\frac{x - 2}{x^{2} - x} - \frac{1}{x^{3} - 3 x^{2} + 2 x}]

\frac{x^{2} - 7 x + 10}{x^{2} - 5 x - 14} \times \frac{x^{3} + 8}{x^{3} - 8} \times \frac{x^{2} + 2 x + 4}{x^{2} - 2 x + 4} \times \frac{x - 7}{x - 5} = ?

y = x^{3} + 2 x^{2} + 7 x + 8

\frac{d y}{d x}

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