Find the expansion of
$(x + 1) (x + 4) (x + 7)$

x^{3} + 12 x^{2} + 39 x + 28

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x^{3} - 12 x^{2} + 39 x + 28

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x^{3} + 12 x^{2} - 39 x + 28

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None of these

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Solution

The correct option is B $x^{3} + 12 x^{2} + 39 x + 28$
$(x + 1) (x + 4) (x + 7)$

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

$= (x^{2} + 5 x + 4) (x + 7)$

$= x^{3} + 7 x^{2} + 5 x^{2} + 35 x + 4 x + 28$

$= x^{3} + 12 x^{2} + 39 x + 28$

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Similar questions

x^{3} - 12 x^{2} + 39 x - 28 = 0

x^{3} - 12 x^{2} + 39 x - 28 = 0

If the roots of the equation $x^{3} - 12 x^{2} + 39 x - 28 = 0$ are in A.P., then find the common difference.

x^{3} - 12 x^{2} + 39 x - 28 = 0

f (x) = x^{3} - 12 x^{2} + 39 x - 28

A P

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