Find the quotient and remainder using polynomial long division. $\newline$ $(2x^{3}-12x^{2}+7x-28)/(2x^{2}+5)$ $\newline$ The quotient is $\newline$ The remainder is

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Q. Find the quotient and remainder using polynomial long division.

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(2x^{3}-12x^{2}+7x-28)/(2x^{2}+5)

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The quotient is

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The remainder is

Perform Division: We will perform polynomial long division to find the quotient and remainder when dividing $(2x^{3}-12x^{2}+7x-28)$ by $(2x^{2}+5)$ . First, we divide the leading term of the dividend, $2x^{3}$ , by the leading term of the divisor, $2x^{2}$ , to find the first term of the quotient. $2x^{3} \div 2x^{2} = x$ .
Find First Term: We multiply the entire divisor $(2x^{2}+5)$ by the term we just found, $x$ , and subtract the result from the dividend. $\newline$ $(2x^{2}+5) \times x = 2x^{3} + 5x$ . $\newline$ Subtract this from the dividend: $\newline$ $(2x^{3}-12x^{2}+7x-28) - (2x^{3} + 5x) = -12x^{2} + 7x - 5x - 28 = -12x^{2} + 2x - 28$ .
Subtract Result: We bring down the next term of the dividend, which is already included in the subtraction result, and repeat the division process. $\newline$ Now we divide the leading term of the new polynomial, $-12x^{2}$ , by the leading term of the divisor, $2x^{2}$ . $\newline$ $-12x^{2} \div 2x^{2} = -6$ .
Repeat Division: We multiply the entire divisor $(2x^{2}+5)$ by the term we just found, $-6$ , and subtract the result from the new polynomial.(\(2x^{ $2$ }+ $5$ ) \times ( $-6$ ) = $-12$ x^{ $2$ } - $30$ \. Subtract this from the new polynomial:(\(-12x^{ $2$ } + $2$ x - $28$ ) - ( $-12$ x^{ $2$ } - $30$ ) = $-12$ x^{ $2$ } + $12$ x^{ $2$ } + $2$ x - $28$ + $30$ = $2$ x + $2$ \.
Find Next Term: Since the degree of the remaining polynomial $2x + 2$ is less than the degree of the divisor $2x^{2}+5$ , we cannot continue the division process. Therefore, $2x + 2$ is the remainder.
Finalize Division: The quotient of the division is the sum of the terms we found: $x - 6$ . The remainder is the last polynomial we obtained: $2x + 2$ .