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What are the quotient and remainder of (3x4 - 2x + 7x-4)= (x - 3)?
a. 3x2 + 772 +21x + 70; 206
c. 3x4 + 7%? +21x2 + 70; 206
b. 3x + 77? + 21x + 70; -206
d. 37° +73+ + 21x + 703 76

Sagot :

MrRoyal

The quotient of dividing (3x⁴ - 2x³ + 7x² - 4) by (x - 3) is 3x³ + 7x² + 28x + 84 and the remainder is 248

How to determine the quotient and the remainder?

The division expression is given as:

(3x⁴ - 2x³ + 7x² - 4) / (x - 3)

Set the divisor to 0

x - 3 = 0

Solve for x

x = 3

Substitute x = 3 in 3x⁴ - 2x³ + 7x² - 4

3x⁴ - 2x³ + 7x² - 4 = 3(3)⁴ - 2(3)³ + 7(3)² - 4

Evaluate

3x⁴ - 2x³ + 7x² - 4 = 248

This means that the remainder is 248

Recall that:

Dividend = Divisor * Quotient + Remainder

So, we have:

3x⁴ - 2x³ + 7x² - 4 = x - 3 * Q + 248

Subtract 248 from both sides

3x⁴ - 2x³ + 7x² - 252 = x - 3 * Q

Divide both sides by x - 3

Q = (3x⁴ - 2x³ + 7x² - 252)/(x - 3)

Factorize the numerator

Q = (x - 3)(3x³ + 7x² + 28x + 84)/(x - 3)

Cancel out the common factors

Q = 3x³ + 7x² + 28x + 84

Hence, the quotient is 3x³ + 7x² + 28x + 84 and the remainder is 248

Read more about polynomial division at:

https://brainly.com/question/25289437

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