Use long division to find the quotient below.
$\left(4 x^3+2 x^2+50\right)-(2 x+5)$
A. $2 x^2-4 x+10$
B. $2 x^2-3 x+10$
C. $2 x^2+3 x+10$
D. $2 x^2+4 x+10$
Answer (1)
Perform polynomial long division of ( 4 x 3 + 2 x 2 + 50 ) by ( 2 x + 5 ) .
Divide 4 x 3 by 2 x to get 2 x 2 , then multiply ( 2 x + 5 ) by 2 x 2 and subtract from the dividend.
Continue the process to find the quotient.
The quotient is 2 x 2 − 4 x + 10 .
Explanation
Understanding the Problem We are asked to perform polynomial long division to find the quotient of ( 4 x 3 + 2 x 2 + 50 ) divided by ( 2 x + 5 ) . This involves dividing the polynomial 4 x 3 + 2 x 2 + 50 by the binomial 2 x + 5 .
Performing Long Division To perform long division, we set up the division as follows:
\t\t\t\t\t 2 x 2 − 4 x + 10 \t\t\t 2 x + 5 | 4 x 3 + 2 x 2 + 0 x + 50 \t\t\t\t\t − ( 4 x 3 + 10 x 2 ) \t\t\t\t\t------------------- \t\t\t\t\t\t\t − 8 x 2 + 0 x \t\t\t\t\t\t\t − ( − 8 x 2 − 20 x ) \t\t\t\t\t\t\t------------------- \t\t\t\t\t\t\t\t\t 20 x + 50 \t\t\t\t\t\t\t\t\t − ( 20 x + 50 ) \t\t\t\t\t\t\t\t\t------------------- \t\t\t\t\t\t\t\t\t\t\t 0
Determining the Quotient
Divide 4 x 3 by 2 x to get 2 x 2 . Multiply ( 2 x + 5 ) by 2 x 2 to get 4 x 3 + 10 x 2 . Subtract this from 4 x 3 + 2 x 2 to get − 8 x 2 .
Bring down the 0 x term. Divide − 8 x 2 by 2 x to get − 4 x . Multiply ( 2 x + 5 ) by − 4 x to get − 8 x 2 − 20 x . Subtract this from − 8 x 2 + 0 x to get 20 x .
Bring down the 50 term. Divide 20 x by 2 x to get 10 . Multiply ( 2 x + 5 ) by 10 to get 20 x + 50 . Subtract this from 20 x + 50 to get 0 .
The quotient is 2 x 2 − 4 x + 10 and the remainder is 0 .
Final Answer Therefore, the quotient is 2 x 2 − 4 x + 10 .
Examples
Polynomial long division is a method used to divide one polynomial by another. It's similar to long division with numbers, but with variables and exponents involved. For example, if you're designing a rectangular garden and know the area and one side's length as polynomials, you can use polynomial long division to find the length of the other side. This technique is also fundamental in simplifying complex algebraic expressions and solving equations in various fields like engineering and computer science.
