BUse long division or an area model to divide polynomials. Is there a remainder? If so, then the linear factors are not factors of the polynomial.
A(x-1), (x+1), (x-7), (x+7)
BSee solution.
Practice makes perfect
Consider the given polynomial.
x^4-6x^3-6x^2+6x-7
This is a fourth degree polynomial which means we will have a maximum of four linear factors. Since the leading coefficient is 1, we can guess the factored form of the given polynomial.
1(x- a)(x- b)(x- c)(x- d)In this case the roots are a, b, c, and d. If we multiply all of these roots we will get the constant term, abcd.
a b c d=- 7
Therefore, a, b, c, and d are factors of - 7. Notice that 7 is a prime number. This means we can only have ± 1 and ± 7 as its factors. This gives us four possible linear factors.
(x-1), (x+1), (x-7), (x+7)
To check if (x+1) is a factor of the given polynomial, we can divide the polynomial by (x+1) and see if the remainder is 0.
x^4-6x^3-6x^2+6x-7/x+1
We can use an area model to help us do it. We will write the terms of the divisor as the headers for the rows in the model and then divide term by term. The first term in the dividend is x^4.
x
x^4
1
To get this term, we need to multiply x+1 by x^3. Let's do it!
x^3
x
x^4
1
x^3
The next term of the dividend is - 6x^3. We already have x^3 as the x^3-term which means we need to consider - 7x^3 in the next step.
x^3
x
x^4
- 7x^3
1
x^3
To get this term, we need to multiply x+1 by - 7x^2. Let's do it!
x^3
- 7x^2
x
x^4
- 7x^3
1
x^3
- 7x^2
The next term of the dividend is - 6x^2. We already have - 7x^2 as the x^2-term which means we need to consider x^2 in the next step.
x^3
- 7x^2
x
x^4
- 7x^3
x^2
1
x^3
- 7x^2
To get this term, we need to multiply x+1 by x. Let's do it!
x^3
- 7x^2
x
x
x^4
- 7x^3
x^2
1
x^3
- 7x^2
x
The next term of the dividend is 6x. We already have x as the x-term which means we need to consider 5x in the next step.
x^3
- 7x^2
x
x
x^4
- 7x^3
x^2
5x
1
x^3
- 7x^2
x
To get this term, we need to multiply x+1 by 5. Let's do it!
x^3
- 7x^2
x
5
x
x^4
- 7x^3
x^2
5x
1
x^3
- 7x^2
x
5
We got 5 as the constant term. The constant term in the original polynomial is - 7. Therefore, we have - 12 as the remainder.
x^3
- 7x^2
x
5
Remainder
x
x^4
- 7x^3
x^2
5x
- 12
1
x^3
- 7x^2
x
5
This means that (x+1) is not a factor of this polynomial. Let's repeat this process for (x-1).
x^3
- 5x^2
- 11x
- 5
Remainder
x
x^4
- 5x^3
- 11x^2
- 5x
- 12
- 1
- x^3
5x^2
11x
5
Once again, we got - 12 as the remainder. This means that (x-1) is not a factor of this polynomial.
