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Algebra 1 : How to divide polynomials

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Example Questions

Example Question #571 : Intermediate Single Variable Algebra

Subtract:

$\frac{2x}{x-1} - \frac{3}{x^{2}-1}$

Possible Answers:

$\frac{2x^{2}-3}{\left ( x+1 \right )\left ( x-1 \right )}$

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

$\frac{2}{x-1}$

$\frac{2x^{2}+2x-3}{\left ( x+1 \right )\left ( x-1 \right )}$

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

Correct answer:

$\frac{2x^{2}+2x-3}{\left ( x+1 \right )\left ( x-1 \right )}$

Explanation:

First let us find a common denominator as follows:

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

Now we can subtract the numerators which gives us : $2x^{2}+2x-3$

So the final answer is $\frac{2x^{2}+2x-3}{\left ( x+1 \right )\left ( x-1 \right )}$

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Example Question #62 : Solving Rational Expressions

Simplify:

$\frac{6x^{2}-7x-3}{6x^{2}+x-15}$

Possible Answers:

$\frac{-\left ( 7+3 \right )}{x+15}$

$\frac{7x}{5}$

$\frac{7}{5}$

$\frac{3x+1}{3x+5}$

None of the above

Correct answer:

$\frac{3x+1}{3x+5}$

Explanation:

Factor both the numerator and the denominator which gives us the following:

$\frac{\left ( 2x-3 \right )\left ( 3x+1 \right )}{\left ( 2x-3 \right )\left ( 3x+5 \right )}$

After cancelling $\left ( 2x-3 \right )$ we get

$\frac{3x+1}{3x+5}$

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Example Question #2 : Multiplication And Division

Simplify the following: $(24x^8-2x^6+12x^2)\div x^2$

Possible Answers:

24x^6-2x^4+12

$24x^{10}-2x^8+12x^4$

12x^6-x^4+6

$24x^{16}-2x^{12}+12x^4$

24x^4-2x^3+12

Correct answer:

24x^6-2x^4+12

Explanation:

We are dividing the polynomial by a monomial. In essence, we are dividing each term of the polynomial by the monomial. First I like to re-write this expression as a fraction. So,

$(24x^8-2x^6+12x^2)\div x^2 = \frac{24x^8-2x^6+12x^2}{x^2}$

So now we see the three terms to be divided on top. We will divide each term by the monomial on the bottom. To show this better, we can rewrite the equation. $\frac{24x^8-2x^6+12x^2}{x^2} = \frac{24x^8}{x^2}-\frac{2x^6}{x^2}+\frac{12x^2}{x^2}$

Now we must remember the rule for dividing variable exponents. The rule is $\frac{x^a}{x^b} = x^{a-b}$ So, we can use this rule and apply it to our expression above. Then, $\frac{24x^8}{x^2}-\frac{2x^6}{x^2}+\frac{12x^2}{x^2} = 24x^{8-2}-2x^{6-2}+2x^{2-2}=24x^6-2x^4+12$

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Example Question #21 : Intermediate Single Variable Algebra

Divide:

$(x^{3} -23x+10) \div (x-5)$

Possible Answers:

$x^{2} - 5x + 16 + \frac{87}{x-5}$

$x^{2}+1+\frac{12}{x-5}$

$x^{2} + 5x + 2$

$x^{2} + 5x + 2 + \frac{20}{x-5}$

$x^{2} - 5x + 16 - \frac{73}{x-5}$

Correct answer:

$x^{2} + 5x + 2 + \frac{20}{x-5}$

Explanation:

First, rewrite this problem so that the missing $x^{2}$ term is replaced by $0x^{2}$

$(x^{3} + 0x^{2} -23x+10) \div (x-5)$

Divide the leading coefficients:

$x^{3} \div x = x^{2}$ , the first term of the quotient

Multiply this term by the divisor, and subtract the product from the dividend:

$x^{2} (x-5) = x^{3} -5x^{2}$

$(x^{3} + 0x^{2} -23x+10) - (x^{3} -5x^{2}) = 5x^{2}-23x+10$

Repeat this process with each difference:

$5x^{2} \div x = 5x$ , the second term of the quotient

$5x (x-5) = 5 x^{2} -25$

$5x^{2}-23x+10 - ( 5x^{2} -25x) = 2x+10$

One more time:

$2x \div x = 2$ , the third term of the quotient

2 (x-5) = 2x-10

2x+10 - (2x-10) = 20 , the remainder

The quotient is $x^{2} + 5x + 2$ and the remainder is ; this can be rewritten as a quotient of

$x^{2} + 5x + 2 + \frac{20}{x-5}$

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Example Question #21 : How To Divide Polynomials

Divide:

$\left (x^{2} + 6x -7 \right )\div (x+3)$

Possible Answers:

$x+ 3 + \frac{16}{x+3}$

$x- 3 - \frac{16}{x+3}$

$x+ 3 - \frac{16}{x+3}$

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

$x+ 3 + \frac{2}{x+3}$

Correct answer:

$x+ 3 - \frac{16}{x+3}$

Explanation:

Divide the leading coefficients to get the first term of the quotient:

$x^{2} \div x = x$ , the first term of the quotient

Multiply this term by the divisor, and subtract the product from the dividend:

$x (x+3) = x^{2} + 3x$

$\left (x^{2} + 6x -7 \right ) - (x^{2}+3x) = 3x -7$

Repeat these steps with the differences until the difference is an integer. As it turns out, we need to repeat only once:

$3x\div x = 3$ , the second term of the quotient

3 (x+3) = 3x + 9

(3x -7) - (3x + 9) = -16 , the remainder

Putting it all together, the quotient can be written as $x+ 3 - \frac{16}{x+3}$ .

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Example Question #22 : How To Divide Polynomials

Divide the polynomials: $\frac{x^2-28x-60}{x^2-3x-10}$

Possible Answers:

-25x-70

$\frac{x-30}{x-5}$

$\frac{x+30}{x-5}$

$\frac{x+30}{x+5}$

Correct answer:

$\frac{x-30}{x-5}$

Explanation:

In order to divide these polynomials, we will need to factorize both the top and the bottom expressions.

$\frac{x^2-28x-60}{x^2-3x-10} = \frac{(x-30)(x+2)}{(x-5)(x+2)}$

Cancel out the common terms in the numerator and denominator.

The answer is: $\frac{x-30}{x-5}$

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Example Question #82 : Variables

Divide the polynomials x^2-2x+1 and 2x^2-2 .

Possible Answers:

$\frac{x-1}{2x-2}$

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

$\frac{x-1}{x-2}$

$\frac{x+1}{2x+2}$

$\textup{The polynomials cannot be factored.}$

Correct answer:

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

Explanation:

Write an expression to divide the polynomials.

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

Both polynomials can be factorized.

The numerator will be simplified to (x-1)^2 .

The denominator cannot be factorized as is. Pull a common factor of two in the denominator.

2x^2-2 = 2(x^2-1)

The x^2-1 term can be factorized to (x+1)(x-1) .

We can now rewrite the fraction.

$\frac{x^2-2x+1}{2x^2-2} = \frac{(x-1)^2}{2(x+1)(x-1)}$

The common (x-1) terms in the numerator and denominator can be cancelled.

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

Distribute the two in the denominator through both terms in the binomial.

The answer is: $\frac{x-1}{2x+2}$

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Algebra 1 : How to divide polynomials

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #571 : Intermediate Single Variable Algebra

Example Question #62 : Solving Rational Expressions

Example Question #2 : Multiplication And Division

Example Question #21 : Intermediate Single Variable Algebra

Example Question #21 : How To Divide Polynomials

Example Question #22 : How To Divide Polynomials

Example Question #82 : Variables

All Algebra 1 Resources

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