Answer:
1.
- (x -3) is a factor of f(x)
- (3x+2) is a factor of f(x)
- f(-1) = 0
- f(x) divided by (x+1) has a remainder of 0
- f(x) = 0 when x = 3
2.
3.
Step-by-step explanation:
Each of these polynomials has a sum of odd-degree coefficients equal to the sum of even-degree coefficients. This means -1 is a root. The cubics can be reduced to quadratics by removing those factors of (x+1).
In any event, I like to use a graphing calculator to show me the roots of higher degree polynomials.
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1. Factoring out (x+1) reduces the cubic to 3x^2 -7x -6, which will have a factor of (x -3). Dividing that out gives the remaining factor of 3x+2.
So, the factorization is f(x) = (x+1)(3x+2)(x-3). It will have zeros at -1, -2/3 and 3. Knowing these things can help you choose the correct true statements from the list.
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2. Factoring out (x+1) reduces the cubic to x^2 -2x -8. Factoring that tells you the factorization is f(x) = (x+2)(x+1)(x-4). This is all you need to know to pick the correct factors from the list.
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3. Factoring out (x-5) reduces the quartic to x^3 +x^2 -x -1. The sum of coefficients is zero, so you know x=1 is a root. Also, the sums of odd-degree and even-degree coefficients are the same (0), so you know that -1 is also a root. Factoring out known roots gives you a remaining factor of (x+1), so the factorization is f(x) = (x-5)(x-1)(x+1)^2.
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<em>Comment on factoring the quadratic in Problem 1</em>
Maybe you have learned to factor quadratics like 3x^2 -7x -6. The job involves finding factors of (3)(-6) = -18 that have a sum of -7. Those factors would be -9 and 2. With this knowledge, you can write the factorization as ...
(3x-9)(3x+2)/3 . . . . . all of the 3s in this expression are copies of the leading coefficient.
The factor (3x-9) is the only one that conveniently divides by 3, so this expression reduces to (x-3)(3x+2).
Of course, you can find the roots of the quadratic however you like if factorization doesn't work well for you.