Dividing by Monomials

You have already found the products of polynomials.

3y(y2 + 2y + 7) = 3y3 + 6y2 + 21y
(x + 3)(x + 2) = x2 + 5x + 6

You can use the same principles to divide polynomials

  • dividing by a monomial 3y³ + 6y² + 21y/3y = y² + 2y + 7
  • dividing by a binomial x² + 5x + 6/x + 3 = X + 2

Two methods for dividing a polynomial by a monomial are factoring and dividing each term.

Example

Simplify:

  1. (6x3 – 8x2 + 12x) ÷ 2x
  2. (25y4 + 30y2 – 20y) ÷ -5y

Solution

  1. 6x3 – 8x2 + 12x
    Look for common factors.
    6, 8, and 12 have the common factor 2. The common factor of x3, x2 and x is x.
    6x3 – 8x2 + 12x = 2x(3x2 – 4x + 6)
    (6x3 – 8x2 + 12x) ÷ 2x = 3x2 – 4x + 6
  2. Divide each term.
    25y4 + 30y² – 20y/-5y = -25y4/-5y + 30y²/-5y + -20y/-5y = -5y3 – 6y + 4

Dividing By Binomials

Two methods for dividing a polynomial by a binomial are factoring and long division.

Example

Simplify:

  1. (x2 – 3x – 4) ÷ (x + 1)
  2. x – 7/2x² – 11x – 21
  3. (6y2 + 7y – 3) ÷ (3y – 1)

Solution

  1. Factor. x² – 3x – 4/x + 1 = (x – 4)(x + 1)/(x + 1) = x – 4
  2. Factor> x – 7/2x² – 11x – 21 = (x – 7)/(x – 7)(2x + 3) = 1/2x + 3
  3. Use the format of long division:

__________2y + 3
3y – 1 )6y² + 7y – 3
_____|6y² – 2y
_____|____9y – 3
__________9y – 3
_____|________0

6y² ÷ 3y = 2y
Multiply 3y – 1 by 2y and subtract. 7y – (-2y) = 9y
9y ÷ 3y = 3
Multiply 3y – 1 by 3 and subtract.

(6y² + 7y – 3) ÷ (3y – 1) = 2y + 3

Complex Fractions

A complex fraction is a fraction that has a fraction in the numerator or denominator. In other words, it is a fraction divided by a fraction. Complex fractions can contain variable expressions. To simplify, multiply by reciprocal, then factor and cancel any common factors.

Example

Simplify:

  1. (x²/9)/(3x/4)
  2. (9y²/7)/12y
  3. x² + 7x/4/x² – 49/x

Solution

Solution

Use the reciprocal and multiply.

  1. /9 / 3x/4 = /9 × 4/3x = 4x/27
  2. 9y²/7/12y = 9y²/7 × 1/12y. Division by 12y is multiplication by its reciprocal 1/12y
    3 × 3 × y × y/7 × 1/3 × 4 × y = 3y/28. Factor, then cancel common factors.
  3. x² + 7x/4/² – 49x = x² + 7x/4 × x/x² – 49. Multiply by the reciprocal.
    x(x + 7)/4 × x/(x + 7)(x – 7) = /4(x – 7). Factor, then cancel common factors.

Adding Polynomial Fractions

When adding (or subtracting) algebraic fractions, follow the same process as adding number fractions. The first step is to write equivalent fractions that have the same (common) denominator.

  1. Find the least common denominator (LCD) of the fractions.
  2. Write equivalent fractions using the LCD.
  3. Add (or subtract) the numerators.
  4. Simplify and reduce the resulting fraction.

A quick method for finding the LCD is to multiply the denominators. But multiplying the denominators often gives a rather large expression. Instead of multiplying, factor the denominators. Use the least common multiple. The LCM will have one of each shared factor and all of the factors that are not shared. (To review LCD and LCM, see the sections on Divisibility and Fractions in the previous chapter.)

Example

Find the least common multiple:

  1. 14s and 6s²
  2. y² – 25 and y² + 6y + 5

Solution

  1. Factor:
    14s = 2 × 7 × s
    6s² = 2 × 3 × s × s
    The LCM of 14s and 6s2 is 2 × 7 × s × 3 × s = 42s²
  2. Factor.
    y² – 25 = (y + 5)(y – 5)
    y² + 6y + 5 = (y + 1)(y + 5)
    The LCM of y² – 25 and y² + 6y + 5 is (y + 5)(y – 5)(y + 1).
Example

Solve: 1/2 + 3/2x + 5/ = 1

Solution

1/2 + 3/2x + 5/ = 1 Multiply both sides by the LCD.
2x²[1/2 + 3/2x + 5/] = 2x² Use the distributive property.
2x²/2 + 6x²/2x + 10x²/ = 2x² Cancel common factors in each fraction.
x² + 3x + 10 = 2x² Subtract 2x² from both sides and multiply by -1.
x² – 3x – 10 = 0 Factor.
(x – 5)(x + 2) = 0 so x = 5 and x = -2

Example

Solve: 5y/y – 2 = 15/y + 10/y – 2

Solution

The least common denominator is y × (y – 2) = y(y – 2).
y(y – 2)[5y/y – 2] = y(y – 2)[15/y + 10/y – 2] Multiply both sides by the LCD.
Cancel common factors.
5y² = 15(y – 2) + 10y Use the distributive property.
5y² = 15y – 30 + 10y Set equations equal to zero.
5y² – 25y + 30 = 0 Divide by 5.
y² – 5y + 6 = 0 Factor.
(y – 2)(y – 3) = 0 so y = 2 and y = 3.

Check the answers by substituting in the original equation. If y = 2, the denominator y – 2 is equal to zero. Division by zero is undefined. So the only solution is y = 3.

How Many Solutions?

The number of solutions for polynomial equations can vary. Some equations have infinite solutions or no solutions. Therefore, you need to be careful when performing operations on an equation that you don’t lose a possible solution or get an extraneous solution, which is a solution that does not satisfy the original equation. The value y = 2 in the Example above is an extraneous solution.

You might lose a solution if you divide both sides by the variable, and you might get an extraneous solution if you square both sides of an equation.

Example

Solve:5x + 5 = 0

Solution

5x + 5 = 0 Subtract 5 from both sides.
5x = -5 It looks like the next step is to square both sides, so x = 5
But on the GMAT, a square root is never a negative number. There are no solutions.

Example

Solve:

  1. 4x³ = x
  2. 16 – y² = 10(4 + y)

Solution

  1. 4x³ = x
    It looks like the way to solve the equation is to divide both sides by x.
    4x³ = x
    4x² = 1
    x² = 1/4
    x = 1/2 and x = -1/2
    But by dividing both sides by x, you lose a value for x.
    4x³ = x
    4x³ – x = 0
    x(4x² – 1) = 0
    x(2x + 1)(2x – 1) = 0
    So x = 0 is another solution. The solutions are x = 0, x = 1/2 and x = -1/2.
  2. 16 – y² = 10(4 + y)
    Factor
    (4 + y)(4 – y) = 10(4 + y)
    It looks like the next step is to divide both sides by 4 + y.
    4 – y = 10, y = 4 – 10 = -6
    But by dividing both sides by 4 + y, you lose a value for y.
    16 – y² = 10(4 + y)
    16 – y² = 40 + 10y
    y² + 10y + 24 = 0
    (y + 6)(y + 4) = 0, y = -6 and y = -4
    So y = -4 is another solution.
Example

Solve:2x = x – 4

Solution

2x = x – 4
Square both sides.
2x = (x – 4)² = x² – 8x + 16
x² – 10x + 16 = 0
(x – 2)(x – 8) = 0
x = 2 and x = 8
Check the answers by substituting in the original equation.
If x = 2, then √2x = x – 4 becomes √4 = 2 – 4 = -2 . So x = 2 is an extraneous solution.
The only solution is x = 8.

Example

Solve: If x/y = 16 and x/y² = 8, what is xy?

Solution

Multiply both sides by y: x/y = 16 → x = 16y
Multiply both sides by y²: x/y² = 8 → x = 8y²
Set the two values of x equal: 16y = 8y²
2y = y² → y² – 2y = 0 → y(y – 2) = 0 → y = 2 and y = 0

You are solving for xy. Substitute the values of y in either equation to find x.

x/y = 16 and y = 2 → x/2 = 16 → x = 32
x/y = 16 and y = 0 → x/0 = 16
Division by zero is undefined, so y = 0 is an extraneous solution.

The only solution is xy = 32 × 2 = 64.

Polynomials and Radicals

You can apply the properties of multiplying and dividing polynomials to simplify equations with radicals. Remember that the simplest form of an expression does not have radicals in the denominator. (To review rules of radicals, see the section on Roots in the previous chapter.)

Example 1

Simplify.

(a) 6/2√3
(b) √x2/2

Solution

To simplify, multiply by a fraction equal to 1 with numerator and denominator equal to the radical.

(a) 6/2√3 = 6/2√3 × √3/√3 = 6√3/2 × 3 = √3

(b)√x2/2 =√x2/√2     First apply the rule of exponents (a/b)n = an/bn

√x2/√2 = |x|/√2 × √2/√2 = |x| � √2 /2

Example

Simplify:

(√x – √y )(√x + √y )

Solution

(√x – √y )(√x + √y ) = x – y              Use (a + b)(a – b) = a2 – b2.

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