The area of a rectangle is . The width of the rectangle is . What is the length?
Even though it does not seem like it, factoring is a form of division. Each factor goes into the larger polynomial evenly, without a remainder. For example, take the polynomial . If we use factoring by grouping, we find that the factors are . If we multiply these three factors together, we will get the original polynomial. So, if we divide by , we should get .
How many times does go into ? times.
Place above the term in the polynomial.
Multiply by both terms in the divisor ( and -3) and place them until their like terms. Subtract from the dividend . Pull down the next two terms and repeat.
goes into 4 times.
After multiplying both terms in the divisor by -4, place that under the terms you brought down. When subtracting we notice that everything cancels out. Therefore, just like we thought, is a factor.
When dividing polynomials, not every divisor will go in evenly to the dividend. If there is a remainder, write it as a fraction over the divisor.
Solution: Set up the problem using a long division bar.
How many times does go into ? times.
Multiply by the divisor. Subtract that from the dividend.
Repeat the previous steps. Now, how many times does go into ? 4 times.
At this point, we are done. cannot go into because it has a higher degree. Therefore, is a remainder. The complete answer would be .
. Determine if goes evenly into . If so, try to factor the divisor and quotient further.
Solution: First, do the long division. If goes in evenly, then the remainder will be zero.
This means that and both go evenly into . Let’s see if we can factor either or further.
Therefore, . You can multiply these to check the work. A binomial with a degree of one is a factor of a larger polynomial, , if it goes evenly into it. In this example, and are all factors of . We can also say that 1, 1, -3, and are all solutions of .
Factor Theorem: A polynomial, , has a factor, , if and only if .
In other words, if is a solution or a zero , then the factor, divides evenly into .
Determine if 5 is a solution of .
Solution: To see if 5 is a solution, we need to divide the factor into . The factor that corresponds with 5 is .
Since there is a remainder, 5 is not a solution.
Intro Problem Revisit
First, do the long division.
This means that and both go evenly into .
can't be factored further, so it is the rectangle's length.
2. Is a factor of ? If so, find any other factors.
3. What are the real-number solutions to #2?
4. Determine if 6 is a solution to .
1. Make sure to put a placeholder in for the term.
The final answer is .
2. Divide into and if the remainder is zero, it is a factor.
is a factor. Let’s see if factors further. Yes, the factors of -27 that add up to -6 are -9 and 3. Therefore, the factors of are , and .
3. The solutions would be -4, 9, and 3; the opposite sign of each factor.
4. To see if 6 is a solution, we need to divide into .
Because the remainder is not zero, 6 is not a solution.
- Long division (of polynomial)
- The process of dividing polynomials where the divisor has two or more terms.
- The polynomial that divides into another polynomial.
- The polynomial that the divisor goes into. The polynomial under the division bar.
- The answer to a division problem.
Divide the following polynomials using long division.
Determine all the real-number solutions to the following polynomials, given one factor.
Determine all the real number solutions to the following polynomials, given one zero.
Find the equation of a polynomial with the given zeros.
- 4, -2, and
- 1, 0, and 3
- -5, -1, and
- Challenge Find two polynomials with the zeros 8, 5, 1, and -1.