If f(x)=x^3-9x^2+11x+21f(x)=x 3 −9x 2 +11x+21 and x-3x−3 is a factor of f(x)f(x), then find all of the zeros of f(x)f(x) algebra
ically.
1 answer:
Answer:
x = -1, 3, and 7
Step-by-step explanation:
Using grouping:
f(x) = x³ − 9x² + 11x + 21
f(x) = x³ − 9x² + 18x − 7x + 21
f(x) = x (x² − 9x + 18) − 7 (x − 3)
f(x) = x (x − 6) (x − 3) − 7 (x − 3)
f(x) = (x² − 6x) (x − 3) − 7 (x − 3)
f(x) = (x² − 6x − 7) (x − 3)
f(x) = (x + 1) (x − 7) (x − 3)
To use long division, see the picture.
The zeros are x = -1, 3, and 7.
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It's a linear function. We need only two points to draw a graph.
We choose any values of x and calculate the value of y.
for x= 0 → y = 3(0) - 1 = 0 - 1 = -1 → (0, -1)
for x = 2 → y = 3(2) - 1 = 6 - 1 = 5 → (2, 5)
for x = 0 → y = 3(0) - 1/3 = 0 - 1/3 = -1/3 → (0, -1/3)
for x = 2 → y = 3(2) - 1/3 = 6 - 1/3 = 5 2/3 → (2, 5 2/3)
