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Core Connections Integrated III, 2015 View details

2. Section 11.2

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Exercise 83 Page 590

a Let's start by simplifying the numerator and denominator of the rational expression.

\frac{2 x ^{3} + 5 x ^{2} - 3 x}{4 x ^{3} - 4 x ^{2} + x}

SplitIntoFactors

Split into factors

\frac{x \cdot 2 x ^{2} + x \cdot 5 x - x \cdot 3}{x \cdot 4 x ^{2} - x \cdot 4 x + x \cdot 1}

FactorOut

Factor out $x$

\frac{x ( 2 x ^{2} + 5 x - 3 )}{x ( 4 x ^{2} - 4 x + 1 )}

ReduceFrac

$\frac{a}{b} = \frac{a / x}{b / x}$

\frac{2 x ^{2} + 5 x - 3}{4 x ^{2} - 4 x + 1}

To continue simplifying, we have to factor the expression in the numerator and denominator.

Factor the Numerator

To do that, we can use a generic rectangle and a diamond problem. We know that $2 x^{2}$ and $3$ goes into the lower left and upper right corner of the generic rectangle.

To fill in the remaining two corners, we need to find two $x -$ terms that sum to $5 x$ and have a product of $- 6 x^{2} .$

Notice that the product is negative. This means one factor must be negative and the other must be positive.

Product	$a x (b x)$	$a x + b x$	Sum	Is Equal To $5 x ?$
$- 6 x^{2}$	$- 3 x (2 x)$	$- 3 x + 2 x$	$- x$	$\times$
$- 6 x^{2}$	$- 2 x (3 x)$	$- 2 x + 3 x$	$x$	$\times$
$- 6 x^{2}$	$- 6 x (1 x)$	$- 6 x + 1 x$	$- 5 x$	$\times$
$- 6 x^{2}$	$- 1 x (6 x)$	$- 1 x + 6 x$	$5 x$	$✓$

When one factor is $- x$ and the other is $6 x$ we have a product of $- 6 x^{2}$ and a sum of $5 x .$ Now we can complete the diamond and generic rectangle.

To factor the left-hand side, we add each side of the generic rectangle and multiply the sums.

2 x^{2} + 5 x - 3 ⇕ (2 x + (- 1)) (x + 3) ⇕ (2 x - 1) (x + 3)

Factor the Denominator

Again, we want to use a generic rectangle and diamond problem. We know that $4 x^{2}$ and $1$ goes into the lower left and upper right corner of the generic rectangle.

To fill in the remaining two corners, we need to find two $x -$ terms that sum to $14 x$ and have a product of $33 x^{2} .$

Notice that the product is positive but the sum is negative. This means both factors must be negative.

Product	$a x (b x)$	$a x + b x$	Sum	Is Equal To $- 4 x ?$
$4 x^{2}$	$- 1 x (- 4 x)$	$- 1 x + (- 4 x)$	$- 5 x$	$\times$
$4 x^{2}$	$- 2 x (- 2 x)$	$- 2 x + (- 2 x)$	$- 4 x$	$✓$

When both factors are $- 2 x$ we have a product of $4 x^{2}$ and a sum of $- 4 x .$ Now we can complete the diamond and generic rectangle.

To factor the left-hand side, we add each side of the generic rectangle and multiply the sums.

4 x^{2} - 4 x + 1 ⇕ (- 2 x + 1) (- 2 x + 1) ⇕ (1 - 2 x) (1 - 2 x)

Continuing the Simplification

If we replace the expressions in the numerator and denominator with the factored expressions, we can continue simplifying.

\frac{( 2 x - 1 ) ( x + 3 )}{( 1 - 2 x ) ( 1 - 2 x )}

FactorOutNegSign

Factor out a minus sign

\frac{( 2 x - 1 ) ( x + 3 )}{( - 1 ) ( - 1 + 2 x ) ( - 1 ) ( - 1 + 2 x )}

CommutativePropMult

Commutative Property of Multiplication

\frac{( 2 x - 1 ) ( x + 3 )}{( - 1 ) ( - 1 ) ( - 1 + 2 x ) ( - 1 + 2 x )}

MultNegNegTwoPar

$(- a) (- b) = a \cdot b$

\frac{( 2 x - 1 ) ( x + 3 )}{( - 1 + 2 x ) ( - 1 + 2 x )}

CommutativePropAdd

Commutative Property of Addition

\frac{( 2 x - 1 ) ( x + 3 )}{( 2 x - 1 ) ( 2 x - 1 )}

ReduceFrac

$\frac{a}{b} = \frac{a / ( 2 x - 1 )}{b / ( 2 x - 1 )}$

\frac{x + 3}{2 x - 1}

b Before we factor anything, we will simplify the product as much as we can.

\frac{3 x ^{2} - 5 x - 2}{2 x ^{2} - 1 1 x + 1 5} \cdot \frac{2 x ^{2} - 5 x}{3 x ^{3} - 5 x ^{2} - 2 x}

▼

Simplify

SplitIntoFactors

Split into factors

\frac{3 x ^{2} - 5 x - 2}{2 x ^{2} - 1 1 x + 1 5} \cdot \frac{x \cdot 2 x - x \cdot 5}{x \cdot 3 x ^{2} - x \cdot 5 x - x \cdot 2}

FactorOut

Factor out $x$

\frac{3 x ^{2} - 5 x - 2}{2 x ^{2} - 1 1 x + 1 5} \cdot \frac{x ( 2 x - 5 )}{x ( 3 x ^{2} - 5 x - 2 )}

ReduceFrac

$\frac{a}{b} = \frac{a / x}{b / x}$

\frac{3 x ^{2} - 5 x - 2}{2 x ^{2} - 1 1 x + 1 5} \cdot \frac{2 x - 5}{3 x ^{2} - 5 x - 2}

MultFrac

Multiply fractions

\frac{( 3 x ^{2} - 5 x - 2 ) ( 2 x - 5 )}{( 2 x ^{2} - 1 1 x + 1 5 ) ( 3 x ^{2} - 5 x - 2 )}

ReduceFrac

$\frac{a}{b} = \frac{a / ( 3 x ^{2} - 5 x - 2 )}{b / ( 3 x ^{2} - 5 x - 2 )}$

\frac{2 x - 5}{2 x ^{2} - 1 1 x + 1 5}

Next, we will factor the denominator. Like in Part A, we will use a generic rectangle and a diamond problem. We know that

2 x^{2}

and

15

goes into the lower left and upper right corner.

To fill in the remaining corners, we must find $x -$ terms that sum to $- 11 x$ and with a product of $30 x^{2} .$

Notice that the product is positive but the sum is negative. Therefore, both factors must be negative.

Product	$a x (b x)$	$a x + b x$	Sum	Is Equal To $- 11 x ?$
$30 x^{2}$	$- x (- 30 x)$	$- x + (- 30 x)$	$- 31 x$	$\times$
$30 x^{2}$	$- 2 x (- 15 x)$	$- 2 x + (- 15 x)$	$- 17 x$	$\times$
$30 x^{2}$	$- 3 x (- 10 x)$	$- 3 x + (- 10 x)$	$- 13 x$	$\times$
$30 x^{2}$	$- 5 x (- 6 x)$	$- 5 x + (- 6 x)$	$- 11 x$	$✓$

As we can see, when one factor is $- 5 x$ and the other is $- 6 x$ we have a product of $30 x^{2}$ and a sum of $- 11 x .$ Now we can complete the diamond and generic rectangle.