Which polynomial below has roots x=3,x=4,x=−1?

Question

BlackTom AI · Accepted Answer

Let’s restate the problem in my own words: we’re looking for a polynomial that has roots at x = 3, x = 4, and x = -1.

Option provided: f(x) = x^3 − 6x^2 + 5x + 12.

To verify, factor the polynomial using the given roots. If 3, 4, and −1 are roots, the polynomial can be written as a monic cubic times the factors corresponding to these roots: (x − 3)(x − 4)(x + 1).

First multiply (x − 3)(x − 4): this gives x^2 − 7x + 12.

Then multiply by (x + 1): (x^2 − 7x + 12)(x + 1) = x^3 − 7x^2 + 12x + x^2 − 7x + 12 = x^3 − 6x^2 + 5x + 12.

Thus the expansion of (x − 3)(x − 4)(x + 1) exactly yields f(x) = x^3 − 6x^2 + 5x + 12, which matches the given polynomial.

From a root-product perspective, the product of the roots (3)(4)(−1) = −12. For a monic cubic x^3 − (sum of roots)x^2 + (sum of pairwise products)x − (product of roots), we have sum of roots = 3 + 4 + (−1) = 6, sum of pairwise products = 3·4 + 3·(−1) + 4·(−1) = 12 − 3 − 4 = 5, and the constant term is −(product) = −(−12) = 12, yielding x^3 − 6x^2 + 5x + 12, which aligns with the given polynomial.

Now, addressing potential alternative choices (even though none are listed): a polynomial with these exact roots must factor into (x − 3)(x − 4)(x + 1) up to multiplication by a nonzero constant. If the leading coefficient were not 1, you’d multiply all coefficients by that constant, but the monic form here matches the given f(x).

In summary, the provided polynomial f(x) = x^3 − 6x^2 + 5x + 12 has exactly the roots x = 3, x = 4, and x = −1, as demonstrated by direct expansion and by root-sum/product relationships.

ALGII141-027 4.8 Practice Problems

Which polynomial below has roots x=3,x=4,x=−1?

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