<h3>3
Answers: A, C and D</h3>
In other words, everything but choice B
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Explanation:
You could use polynomial long division or synthetic division to check each answer. The shorter route of the two options is synthetic division
There's a much faster way that doesn't involve complicated division. Recall that if (x-k) is a factor of p(x), then p(k) = 0. This is a special case of the remainder theorem.
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Here's a fairly short proof:
Consider a polynomial q(x) such that
p(x) = (x-k)q(x)
which shows that (x-k) is a factor of p(x). We don't need to worry about what q(x) actually is since it will go away effectively.
If we plug x = k into the p(x) function, we get
p(x) = (x-k)q(x)
p(k) = (k-k)q(k)
p(k) = 0*q(k)
p(k) = 0
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How is this useful? Well we can note that the factor (x-3) is in the form (x-k) where k = 3.
So the idea is to plug x = 3 into each of the four functions and see which result in 0. If we get 0 as an output, then we have a factor. Otherwise, it's not a factor.
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Plug x = 3 into the first function
A(x) = x^3 - 2x^2 - 4x + 3
A(3) = 3^3 - 2(3)^2 - 4(3) + 3
A(3) = 0
This shows (x-3) is a factor of the function A(x)
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Repeat for the second function
B(x) = x^3 + 3x^2 - 2x - 6
B(3) = 3^3 + 3(3)^2 - 2(3) - 6
B(3) = 42
We don't get an output of 0, so (x-3) cannot be a factor of B(x)
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Now onto the third function
C(x) = x^4 - 2x^3 - 27
C(3) = 3^4 - 2(3)^3 - 27
C(3) = 0
So (x-3) is a factor of C(x)
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Finally the last function
D(x) = x^4 - 20x - 21
D(3) = 3^4 - 20(3) - 21
D(3) = 0
Therefore (x-3) is a factor of D(x) as well.
