Find the quotient and the remainder when 2x^2 | vedclass

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Question

(A) To divide $2x^2 - 7x - 15$ by $2x - 3$,we use long division:$1$. Divide the first term of the dividend $(2x^2)$ by the first term of the divisor $

Vedclass · Accepted Answer

Quotient $= x - 2$,Remainder $= -21$

Vedclass · Answer

(A) To divide $2x^2 - 7x - 15$ by $2x - 3$,we use long division:$1$. Divide the first term of the dividend $(2x^2)$ by the first term of the divisor $(2x)$: $2x^2 / 2x = x$. This is the first term of the quotient.$2$. Multiply the divisor $(2x - 3)$ by $x$: $x(2x - 3) = 2x^2 - 3x$.$3$. Subtract this from the dividend: $(2x^2 - 7x - 15) - (2x^2 - 3x) = -4x - 15$.$4$. Divide the first term of the new expression $(-4x)$ by the first term of the divisor $(2x)$: $-4x / 2x = -2$. This is the second term of the quotient.$5$. Multiply the divisor $(2x - 3)$ by $-2$: $-2(2x - 3) = -4x + 6$.$6$. Subtract this from the current expression: $(-4x - 15) - (-4x + 6) = -4x - 15 + 4x - 6 = -21$.Thus,the quotient is $x - 2$ and the remainder is $-21$.

Find the quotient and the remainder when $2x^2 - 7x - 15$ is divided by $2x - 3$.

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