Asked by Tameka
what are the roots for x^3-2x^2-7x-4?
Answers
Answered by
Henry
Y = x^3 -2x^2 - 7x - 4 = 0.
By trial and error, I found that (-1)
satisfies the Eq, and it is our 1st root.
x = -1, x + 1 = 0.
Using synthetic(Long-hand) division,
divide the cubic Eq by x+1:
(X^3 - 2x^2 - 7x - 4)/(x+1)=x^2 - 3x - 4
x^3 - 2x^2 -7x - 4 = (x+1)(x^2-3x-4).
Factor the quadratic Eq:
(x+1)(x+1)(x-4) = 0.
x+1 = 0, X = -1.
x-4 = 0, X = 4.
The roots are: -1, and 4.
Use same procedure for remaining problems.
Note: Instead of factoring the Quadratic Eq, you can divide it by x+1
and get x-4.
x-4 = 0, X = 4.
By trial and error, I found that (-1)
satisfies the Eq, and it is our 1st root.
x = -1, x + 1 = 0.
Using synthetic(Long-hand) division,
divide the cubic Eq by x+1:
(X^3 - 2x^2 - 7x - 4)/(x+1)=x^2 - 3x - 4
x^3 - 2x^2 -7x - 4 = (x+1)(x^2-3x-4).
Factor the quadratic Eq:
(x+1)(x+1)(x-4) = 0.
x+1 = 0, X = -1.
x-4 = 0, X = 4.
The roots are: -1, and 4.
Use same procedure for remaining problems.
Note: Instead of factoring the Quadratic Eq, you can divide it by x+1
and get x-4.
x-4 = 0, X = 4.
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