Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
3. Dividing Polynomials
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Exercise 39 Page 178

Consider polynomial long division. to find the missing dimension.

2x+5

Practice makes perfect

The volume of the rectangular prism given below is V=2x^3+17x^2+46x+40.

The volume of a rectangular prism is the product of the dimensions of its length, width, and height. Since we do not know its length, we can write an equation that represents its volume as shown below. l(x+4)(x+2)=2x^3+17x^2+46x+40 Let's first isolate l in the equation.
l(x+4)(x+2)=2x^3+17x^2+46x+40
Simplify left-hand side
l(x(x+4)+2(x+4))=2x^3+17x^2+46x+40
l(x^2+4x+2(x+4))=2x^3+17x^2+46x+40
l(x^2+4x+2x+8)=2x^3+17x^2+46x+40
l(x^2+6x+8)=2x^3+17x^2+46x+40
l=2x^3+17x^2+46x+40/x^2+6x+8
We are left with a division of polynomials. To perform the division, we will use polynomial long division. To do that, all the terms of the dividend, or the numerator, must be present. Let's consider the polynomial in the numerator again. 2x^3+17x^2+46x+40 Since there are no missing terms, we do not need to rewrite the polynomial. Let's divide!
l r x^2+6x+8 & |l 2x^3+17x^2+46x+40
Divide

2x^3/x^2= 2x

r 2x r x^2+6x+8 & |l 2x^3+17x^2+46x+40

Multiply term by divisor

r 2x rl x^2+6x+8 & |l 2x^3+17x^2+46x+40 & 2x^3+12x^2+16x

Subtract down

r 2x r x^2+6x+8 & |l 5x^2 + 30x + 40
Divide

5x^2/x^2= 5

r 2x+5 r x^2+6x+8 & |l 5x^2 + 30x + 40

Multiply term by divisor

r2x + 5 rl x^2+6x+8 & |l 5x^2 + 30x + 40 & 5x^2 + 30x + 40

Subtract down

r 2x+5 r x^2+6x+8 & |l 0
Therefore, the missing dimension is 2x+5.

Checking Our Answer

Checking Our Answer
We can always check our answer to a division exercise by multiplying the divisor and the quotient. If we end with the original dividend, then we know we have solved the exercise correctly. In this case, we would multiply x^2+6x+8 and 2x+5. Let's do it!
(x^2+6x+8)(2x+5)
2x( x^2+6x+8)+5( x^2+6x+8)
Simplify
2x^3+12x^2+16x+5(x^2+6x+8)
2x^3+12x^2+16x+5x^2+30x+40
2x^3+12x^2+5x^2+16x+30x+40
2x^3+17x^2+46x+40
We ended with the same polynomial as the original dividend, so we know our answer is correct!