Claudia wants to test if the binomial 3x−6 is a factor of P(x)=3x^3−9x^2+9x−6.
First, she evaluates P(_[blank 1]_). After simplifying correctly, she is left with a remainder of _[blank 2]_. Based on the remainder, she correctly concludes that 3x−6 _[blank 3]_ a factor of P(x).
i need help filling in the blanks
thank you
No. The binomial 3x-6 would not be a factor. The factored form of P(x)=3x^3−9x^2+9x−6 is P(x)=3(x-2)(x^2-x+1)
To test if the binomial 3x − 6 is a factor of P(x) = 3x^3 − 9x^2 + 9x − 6, Claudia can use polynomial long division. Here's how she can fill in the blanks:
1. In the blank 1, she will substitute the binomial 3x − 6 into P(x), replacing the variable x. So, she evaluates P(2) since 2 represents the blank 1. It would result in 3(2)^3 − 9(2)^2 + 9(2) − 6. Now, simplify the expression.
2. After simplifying the expression, Claudia will have a remainder. She should fill in blank 2 with the correct value of the remainder.
3. Based on the remainder value, Claudia can determine if 3x − 6 is a factor of P(x). If the remainder is 0, then 3x − 6 is a factor of P(x). If the remainder is nonzero, it means that 3x − 6 is not a factor of P(x), and Claudia should fill in blank 3 with the word "is not."
Remember, polynomial long division is used to divide one polynomial by another and determine if the second polynomial is a factor of the first one.
(3x^3−9x^2+9x−6) / (3x-6)
all the 3's cancel
(x^3 - 3x^2 + 3x - 2) / (x - 2)
remainder is zero
3x-6 is a factor