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Third-degree, with zeros of −3, −2, and 1, and passes through the point (3,9).
Found 2 solutions by josgarithmetic, MathTherapy:Answer by josgarithmetic(37585)
Answer by MathTherapy(10131) (Show Source):
You can put this solution on YOUR website!
Construct a polynomial function with the stated properties. Reduce all fractions to lowest terms.
Third-degree, with zeros of −3, −2, and 1, and
passes through the point (3,9).
The general 3rd degree polynomial is f(x) = ax3 + bx2 + cx + d
Or in factored form, it would be f(x) = a(x-r)(x-s)(x-t), where r, s and t are roots of the polynomial equation.
So you have f(x) = a(x+3)(x+1)(x-2)
Use the point (1,11) to plug in and solve for a.
f(1) = 11 = a(1+3)(1+1)(1-2)
11 = a(4)(2)(-1)
11 = a(-8)
a = -11/8
Thus f(x) = -(11/8)(x+3)(x+1)(x-2) = -(11/8)(x+3)(x2 - x -2)
f(x) = -(11/8)(x3 - x2 - 2x + 3x2 - 3x + 6) = -(11/8)(x3 + 2x2 - 5x + 6)
f(x) = -11x3 /8 - 11x2 /4 + 55x/8 - 33/4
Pretty lost on this.
Pretty lost on this. If \(\displaystyle x = \alpha\) is a
zero then \(\displaystyle (x - \alpha)\) is a factor. So \(\displaystyle y = k(x + 2)(x + 1)(x - 3)\).
"Construct a polynomial function with the stated properties. Reduce all fractions to lowest terms.:
Third-degree, with zeros of -2,-1, and 3, and passes through the poing (4,12)"
(Happy)
"Construct a polynomial function with the stated properties. Reduce all fractions to lowest terms.:
Third-degree, with zeros of -2,-1, and 3, and passes through the poing (4,12)"
(Happy)
Now substitute (4, 12) and solve for k.
Vanessa J. Construct a polynomial function with the stated properties. Reduce all fractions to lowest terms. Third-degree, with zeros of −2, −1, and 3, and passes through the point (4,6
1 Expert Answer
Scott B. answered • 04/06/22
Education focused Physics Professor
Let's start with the zeros part.
If a polynomial p(x) has a zero at some particular value like 3, it means that the binomial (x-3) is one of its factors. Meaning, we can write it as p(x)=(x-3)g(x), where g(x) is some further polynomial expression. Since we're actually given three zeros here, we can write p(x)=(x+2)(x+1)(x-3)g(x), where again, g(x) is some further polynomial expression we have yet to find. You can check that x=-2, x=-1, and x=3 all cause this expression to be 0.
Now we use the fact that we're looking for a third degree polynomial. This tells us that, in fact, g(x) is at most some constant (that I'll now relabel a), because (x+2)(x+1)(x-3) is ALREADY a third degree polynomial. If g(x) contains any factors of x, it will raise us to a fourth order polynomial or higher. So, p(x)=a(x+2)(x+1)(x-3), where a is some constant.
To find what a is, we use the last given information, that the curve must pass through the point (4,6). Substituting 4 in for x, we get
p(4)=a(4+2)(4+1)(4-3)=a(6)(5)(1)=30a=6
So, a=6/30, or 1/5
In total, we have
p(x)=(1/5)(x+2)(x+1)(x-3)
You may expand this out if you prefer.
p(x)=(1/5) x3 - (7/5) x - 6/5
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