Renee is asked to find a factor of x^3−5x^2+6x, given that x is one of the linear factors. Which of the following is another correct factor?(1 point)
x−2
x+3
x+2
x^2+5x+6
The factor theorem states that if a polynomial f(x) has a factor (x-a), then f(a) = 0.
So, plugging in x=-2 into the given polynomial,
(-2)^3 - 5(-2)^2 + 6(-2) = -8 - 20 - 12 = -40.
Therefore, x + 2 is NOT a factor of the given polynomial.
Similarly, plugging in x = -3 into the given polynomial,
(-3)^3 - 5(-3)^2 + 6(-3) = -27 - 45 - 18 = -90.
Therefore, x + 3 is NOT a factor of the given polynomial.
Now, plugging in x=2 into the given polynomial,
(2)^3 - 5(2)^2 + 6(2) = 8 - 20 + 12 = 0.
Therefore, x - 2 is a factor of the given polynomial.
Finally, (x^3-5x^2+6x) / (x-2) = x^2 - 3x.
So, x^2 - 3x is another correct factor.
Well, let me take a stab at this!
Let's Factorize the expression x^3 - 5x^2 + 6x.
So, Renee is asked to find a factor of this expression, given that x is one of the linear factors.
Let's try substituting x = 2 into the expression and see what happens.
(2)^3 - 5(2)^2 + 6(2) = 8 - 5(4) + 12 = 8 - 20 + 12 = 0.
So, x = 2 is a factor of x^3 - 5x^2 + 6x.
That means the correct factor is x - 2.
So, the answer is x - 2. Hope that helps!
To find the factor of x^3-5x^2+6x, given that x is one of the linear factors, we can use synthetic division.
Let's test each option by performing synthetic division with x as the assumed factor:
1) For x-2:
2 | 1 - 5 + 6
| 2 - 6
|_____________
1 - 3 + 0
The remainder is zero, so x-2 is a factor.
2) For x+3:
-3 | 1 - 5 + 6
| -3 24
|_____________
1 - 8 + 30
The remainder is nonzero, so x+3 is not a factor.
3) For x+2:
-2 | 1 - 5 + 6
| -2 14
|_____________
1 - 7 + 20
The remainder is nonzero, so x+2 is not a factor.
4) For x^2+5x+6:
This is a quadratic expression, not a linear factor.
Therefore, the correct factor is x-2.
To find a factor of a polynomial, we need to determine values for x that make the polynomial equal to zero. Given that x is a linear factor, we can use synthetic division to divide the polynomial by x−a, where 'a' is a possible value for x. If the remainder is zero, then x−a is indeed a factor of the polynomial.
Let's start by applying synthetic division using the given options.
Option 1: x−2
We divide x^3−5x^2+6x by x−2 using synthetic division:
2 │ 1 -5 6
────
-2 -6
────
1 -7 0
The remainder is zero (the last value in the result is 0), so x−2 is a factor of the polynomial.
Option 2: x+3
We divide x^3−5x^2+6x by x+3:
-3 │ 1 -5 6
────
-3 24
────
1 -8 30
The remainder is not zero, so x+3 is not a factor of the polynomial.
Option 3: x+2
We divide x^3−5x^2+6x by x+2:
-2 │ 1 -5 6
────
-2 14
────
1 -7 20
The remainder is not zero, so x+2 is not a factor of the polynomial.
Option 4: x^2+5x+6
It is already in the form (x−a), so we don't need to perform synthetic division.
Therefore, the correct factor among the given options is x−2.