Use long division to divide. Specify the quotient and the remainder.
[tex](4 x^2+37 x+63) \div(x+7)[/tex]
quotient $\square$
remainder $\square$
Answer (2)
The result of dividing the polynomial 4 x 2 + 37 x + 63 by x + 7 using long division gives a quotient of 4 x + 9 and a remainder of 0 .
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Set up the long division problem with the dividend 4 x 2 + 37 x + 63 and the divisor x + 7 .
Divide the first term of the dividend by the first term of the divisor to get the first term of the quotient: 4 x .
Multiply the divisor by the first term of the quotient and subtract the result from the dividend to get a new dividend: 9 x + 63 .
Divide the first term of the new dividend by the first term of the divisor to get the second term of the quotient: 9 . The quotient is 4 x + 9 and the remainder is 0 , so the final answer is 4 x + 9 .
Explanation
Understanding the Problem We are asked to divide the polynomial 4 x 2 + 37 x + 63 by the binomial x + 7 using long division. Our goal is to find the quotient and the remainder of this division.
Setting up Long Division We set up the long division as follows:
\t\t\t\t\t\t\t\t x + 7 ) 4 x 2 + 37 x + 63
Finding the First Term of the Quotient First, we divide the leading term of the dividend, 4 x 2 , by the leading term of the divisor, x . This gives us 4 x . This is the first term of the quotient. We write 4 x above the 37 x term in the dividend.
\t\t\t\t\t\t\t\t\t 4 x \t\t\t\t\t\t\t\t x + 7 ) 4 x 2 + 37 x + 63
Multiplying the Divisor by the First Term of the Quotient Next, we multiply the divisor, x + 7 , by 4 x to get 4 x ( x + 7 ) = 4 x 2 + 28 x . We write this below the dividend.
\t\t\t\t\t\t\t\t\t 4 x \t\t\t\t\t\t\t\t x + 7 ) 4 x 2 + 37 x + 63 \t\t\t\t\t\t\t\t\t 4 x 2 + 28 x
Subtracting and Bringing Down Now, we subtract 4 x 2 + 28 x from 4 x 2 + 37 x + 63 to get ( 4 x 2 + 37 x + 63 ) − ( 4 x 2 + 28 x ) = 9 x + 63 . We bring this down as the new dividend.
\t\t\t\t\t\t\t\t\t 4 x \t\t\t\t\t\t\t\t x + 7 ) 4 x 2 + 37 x + 63 \t\t\t\t\t\t\t\t\t 4 x 2 + 28 x \t\t\t\t\t\t\t\t\t\t\t\t 9 x + 63
Finding the Second Term of the Quotient Next, we divide the leading term of the new dividend, 9 x , by the leading term of the divisor, x . This gives us 9 . This is the second term of the quotient. We write + 9 next to 4 x above.
\t\t\t\t\t\t\t\t\t 4 x + 9 \t\t\t\t\t\t\t\t x + 7 ) 4 x 2 + 37 x + 63 \t\t\t\t\t\t\t\t\t 4 x 2 + 28 x \t\t\t\t\t\t\t\t\t\t\t\t 9 x + 63
Multiplying the Divisor by the Second Term of the Quotient Now, we multiply the divisor, x + 7 , by 9 to get 9 ( x + 7 ) = 9 x + 63 . We write this below the 9 x + 63 term.
\t\t\t\t\t\t\t\t\t 4 x + 9 \t\t\t\t\t\t\t\t x + 7 ) 4 x 2 + 37 x + 63 \t\t\t\t\t\t\t\t\t 4 x 2 + 28 x \t\t\t\t\t\t\t\t\t\t\t\t 9 x + 63 \t\t\t\t\t\t\t\t\t\t\t\t 9 x + 63
Subtracting to Find the Remainder Finally, we subtract 9 x + 63 from 9 x + 63 to get ( 9 x + 63 ) − ( 9 x + 63 ) = 0 . This is the remainder.
\t\t\t\t\t\t\t\t\t 4 x + 9 \t\t\t\t\t\t\t\t x + 7 ) 4 x 2 + 37 x + 63 \t\t\t\t\t\t\t\t\t 4 x 2 + 28 x \t\t\t\t\t\t\t\t\t\t\t\t 9 x + 63 \t\t\t\t\t\t\t\t\t\t\t\t 9 x + 63 \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t 0
Final Answer Therefore, the quotient is 4 x + 9 and the remainder is 0 .
Examples
Polynomial division is a fundamental concept in algebra with numerous real-world applications. For instance, engineers use polynomial division to model and analyze systems, such as electrical circuits or control systems. By representing system behavior with polynomials, engineers can use division to simplify complex models, identify key components, and predict system responses. This technique is also crucial in cryptography, where polynomial division is used in error-correcting codes to ensure data integrity during transmission. Understanding polynomial division provides a powerful tool for solving a wide range of practical problems in science and engineering.
