Multiply $(7 x - 4 x^{2} + 2 x^{3} - 5)$ with $(3 x - 2)$ .

$6 x^{4} - 16 x^{3} + 29 x^{2} - x + 1$

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$5 x^{4} - 16 x^{3} + 29 x^{2} - x + 10$

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$6 x^{4} - 16 x^{3} + 27 x^{2} - x + 10$

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$6 x^{4} - 16 x^{3} + 29 x^{2} - 29 x + 10$

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Solution

The correct option is D
$6 x^{4} - 16 x^{3} + 29 x^{2} - 29 x + 10$

$(7 x - 4 x^{2} + 2 x^{3} - 5) \times (3 x - 2) = 3 x (7 x - 4 x^{2} + 2 x^{3} - 5) - 2 (7 x - 4 x^{2} + 2 x^{3} - 5)$
= $21 x^{2} - 12 x^{3} + 6 x^{4} - 15 x - 14 x + 8 x^{2} - 4 x^{3} + 10$
= $6 x^{4} - 16 x^{3} + 29 x^{2} - 29 x + 10$

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Similar questions

\lim_{x \to 4} \frac{x^{2} - 7 x + 12}{x^{2} - 3 x - 4}

The expression obtained by multiplying $(7 x - 4 x^{2} + 2 x^{3} - 5)$ with $(3 x - 2)$ is:

The product of $(7 x - 4 x^{2} + 2 x^{3} - 5)$ and $(3 x - 2)$ is

7 x^{2} - 5 x + 2

4 x^{2} - 3 x + 2

9 x^{\frac{2}{3}} - 3 x + 4

u^{\frac{- 1}{2}} + 3 u + 2

\frac{5}{3} x^{3} + 7 x + 16

\frac{4}{3} x^{7} - 3 x^{5} + 2 x^{3} + 1

y = x^{4} - 2 x^{3} + 4 x^{2} - 3 x + 5,

x = \pm 2

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