Question 92556This question is from textbook Algebra and Trigonometry
: Find the polynomial f(x) of degree three that has zeroes at 1,2, and 4 such that f(0)= -16.
This question is from textbook Algebra and Trigonometry
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Being a zero of a polynomial means when you assign a value to x in the polynomial, the polynomial
will have a value of zero.
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Suppose we could form a polynomial that has three values of x for which the polynomial will
have a value of zero. A way that we could do that is to make the polynomial be the product of
three factors ... the factors having the form (x + a), (x + b), and (x + c). The polynomial
would be the product of these three factors. In other words, the polynomial would be:
.
(x + a)*(x + b)*(x + c)
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But think about this ... the value of the polynomial would be zero if any one of the factors
were zero because zero times the other two factors makes the entire polynomial zero.
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We were given three values of x that will make the polynomial go to zero. The first value
was x = 1. Let's use the first factor (x + a) for this. If x equals 1 in this factor
and we want that factor to go to be able to go to zero, then a must equal -1. So our first
factor is (x - 1). Similarly, we want the second factor to go to zero if x = 2. For it to
go to zero when x = 2, then b must equal -2. So our second factor is (x - 2). Similarly,
the third factor needs to go to zero when x is equal to 4. For this to happen, c must be
equal to -4, so the third factor is (x - 4).
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We can now get these three factors in polynomial form by multiplying them together:
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(x - 1)*(x - 2)*(x - 4)
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You can do that multiplication in two stages. First multiply (x - 1) times (x - 2) and then
multiply the resulting product by (x - 4) to get the answer. Assuming you can do the multiplication,
here's the two results for you to use as checks:
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(x - 1)*(x - 2) = x^2 - 3x + 2
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Then doing the second stage multiplication:
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(x^2 - 3x + 2)*(x - 4) = x^3 - 7x^2 + 14x - 8
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You can check this by letting x = 1 first and this polynomial becomes:
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1^3 - 7(1^2) + 14(1) - 8 = 1 - 7 + 14 - 8 = 0
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Next let x = 2 and the polynomial becomes
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2^3 - 7(2^2) + 14(2) - 8 = 8 - 28 + 28 - 8 = 0
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Finally, let x = 4 and the polynomial becomes:
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4^3 - 7(4^2) + 14(4) - 8 = 64 - 112 + 56 - 8 = 0
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So our three zeros work out OK. But we still have another criterion to meet. That criterion
says that when x equals zero, the polynomial must equal -16. Well, our polynomial
does not meet that. Our polynomial is
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x^3 - 7x^2 + 14x - 8
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and when x is equal to zero, all the terms containing x disappear so we are just left with
a value of -8.
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But what happens if we double our polynomial ... multiply all of its terms by 2. That would
mean that our polynomial would be formed from the factors:
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2*(x - 1)*(x - 2)*(x - 4)
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It would still go to zero if x = 1 or if x = 2 or if x = 4, so that property remains unchanged.
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But we already know that (x - 1)*(x - 2)*(x - 4) = x^3 - 7x^2 + 14x - 8
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And if we multiply this by 2 the resulting polynomial is:
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2x^3 - 14x^2 + 28x - 16
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Since this still contains the factors (x - 1), (x - 2), and (x - 4) so it goes to zero
when x = 1 or when x = 2 or when x = 4. But what about when x = 0? In the polynomial
.
2x^3 - 14x^2 + 28x - 16
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if x is set to zero then all the terms containing x are zero and we are left with the
value of the polynomial being -16, just as the problem requires it should be.
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So the answer is that the polynomial is:
.
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We have already checked when x equals zero to find that f(0) = -16. You can further
check the value of f(1), f(2), and f(4) to make sure all three of them are zero.
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Hope this helps you to understand the problem and how it can be solved.
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