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3 years ago

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Mathematics

1 answer:

iren2701 [21]3 years ago

8 0

Answer:

Step-by-step explanation:

Given:

A. $2x^3-x^{2} -6x$

B. $2x^3+8x+4$

C. $3x^4+x^{2} +x-7$

D. $3x^4-3x^{2} +5x-7$

Now, let us evaluate the given expressions one by one.

$(4x^3-4+ 7x)-(2x^3-x-8)\\\Rightarrow 4x^3-4+ 7x-2x^3+x+8\\\Rightarrow 2x^3+8x+ 4$

It is equation B.

So, $(4x^3-4+ 7x)-(2x^3-x-8)$ is equivalent to B.

$(-3x^2+x^4+x)+(2x^4-7+4x)\\\Rightarrow -3x^2+x^4+x+2x^4-7+4x\\\Rightarrow3x^4-3x^{2} +5x-7$

It is equation D.

So, $(-3x^2+x^4+x)+(2x^4-7+4x)$ is equivalent to D.

$(x^{2} -2x)(2x+3)\\\Rightarrow 2x^3-4x^{2} +3x^{2} -6x\\\Rightarrow 2x^3-x^{2} -6x$

It is equation A.

So, $(x^{2} -2x)(2x+3)$ is equivalent to A.

So, answer is:

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