(7x^3 - 17x^2 - 138x - 72) / (7x+4)

Answer 1

To divide the polynomial (7x^3 - 17x^2 - 138x - 72) by (7x + 4), we can use polynomial long division.

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(7x + 4) | 7x^3 - 17x^2 - 138x - 72

Let's start by dividing the first term of the dividend by the first term of the divisor:

7x^3 / 7x = x^2

Now, multiply x^2 by (7x + 4):

x^2 * (7x + 4) = 7x^3 + 4x^2

Subtract this from the original polynomial:

(7x^3 - 17x^2 - 138x - 72) - (7x^3 + 4x^2) = -21x^2 - 138x - 72

Now, bring down the next term:

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(7x + 4) | 7x^3 - 17x^2 - 138x - 72
-21x^2
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-38x - 72

Now, divide the first term of what remains by the first term of the divisor:

-38x / 7x = -5

Multiply -5 by (7x + 4):

-5(7x + 4) = -35x - 20

Subtract this from the current dividend:

(-38x - 72) - (-35x - 20) = -3x - 52

We now bring down the last term:

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(7x + 4) | 7x^3 - 17x^2 - 138x - 72
-21x^2 - 5
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-38x - 72
-38x - 56
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-16

Since the degree of the remainder is less than the degree of the divisor, we have reached the end of the division.

Therefore, the quotient is x^2 - 5 and the remainder is -16.

Thus, (7x^3 - 17x^2 - 138x - 72) / (7x + 4) = x^2 - 5 + (-16)/(7x + 4)