Question 1189262: a) Use the remainder theorem to find the remainder when 𝑓(𝑥)=−4𝑥^3 +5𝑥^2 + 8 is
divided by 𝑥+3. Then use the Factor Theorem to determine whether 𝑥 + 3 is a
factor of 𝑓(𝑥).
b) Use the Rational Zeros Theorem to find all the real zeros of the polynomial
function 𝑓(𝑥)=x^4 -x^3 -6x^2 +4x +8
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
Remainder theorem: Dividing p(x) over (x-k) yields some quotient q(x) and a remainder p(k).
Special case: if the remainder is 0, then (x-k) is a factor of p(x).
Compare (x+3) with (x-k) to see that k = -3
It might help to rewrite x+3 as x-(-3).
f(x) = -4x^3+5x^2+8
f(-3) = -4(-3)^3+5(-3)^2+8
f(-3) = 161
Answers:
The remainder is 161
(x+3) is not a factor of f(x) because we didn't get a remainder of 0.
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Part (b)
Due to the leading coefficient being 1, this means we simply need to list the plus/minus version of each factor of the last term 8
The list of all possible rational roots:
1, -1, 2, -2, 4, -4, 8, -8
Then check each value to see if we get f(x) = 0 or not.
If we tried x = 1, then,
f(x) = x^4-x^3-6x^2+4x+8
f(1) = (1)^4-(1)^3-6(1)^2+4(1)+8
f(1) = 6
We don't get a result of 0, which means x = 1 is not a root or zero of the function.
Trying x = -1 leads to:
f(x) = x^4-x^3-6x^2+4x+8
f(-1) = (-1)^4-(-1)^3-6(-1)^2+4(-1)+8
f(-1) = 0
We get zero this time, so x = -1 is a root of f(x)
Repeat these steps for the remaining possible roots listed above. You should find that only the following are roots: -2, -1, 2
Side note: x = 2 is a double root
Answer: -2, -1, and 2 are zeros of f(x)
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