we're learning how to solve polynomial functions and i have no idea what im doing so if its easier to start with r(x) = x^4-6x^3+12x^2+6x-13. i must find all the zeros of that function. using synthetic substitution.
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Using the rational root theorem, we try possible factors (-1 , 1 , -13 , 13) and we can see that x = -1 is a solution
(-1)^3 - 5 * (-1)^2 + 7 * (-1) + 13 =>
-1 - 5 - 7 + 13 =>
-13 + 13 =>
0
So, (x + 1) is a factor
(x^3 - 5x^2 + 7x + 13) / (x + 1)
Now, when I divide on here, I don't do the whole synthetic division thing that they teach you, namely because Yahoo!Answers formats in a funny way that I'm not too comfortable with. So, what I do is follow these steps:
1) Divide the highest powered term in the numerator by the highest powered term in the denominator
2) Multiply the number I got in step 1 with the denominator
3) Subtract that from the numerator
4) Repeat steps 1 through 3, until I can't do it anymore
x^3 / x = x^2
x^2 * (x + 1) = x^3 + x^2
x^3 - 5x^2 - (x^3 + x^2) => x^3 - 5x^2 - x^3 - x^2 = -6x^2
-6x^2 / x = -6x
-6x * (x + 1) = -6x^2 - 6x
-6x^2 + 7x - (-6x^2 - 6x) => -6x^2 + 7x + 6x + 6x = 13x
13x / x = 13
13 * (x + 1) = 13x + 13
13x + 13 - (13x + 13) = 13x + 13 - 13x - 13 = 0
(x + 1) * (x^2 - 6x + 13) = (x^3 - 5x^2 + 7x + 13)
Now we can use the quadratic formula to find the other values for x
x^2 - 6x + 13 = 0
x = (6 +/- sqrt(36 - 52)) / 2
x = (6 +/- sqrt(-16)) / 2
x = (6 +/- 4i) / 2
x = 3 +/- 2i
x = -1 , 3 + 2i , 3 - 2i
(-1)^3 - 5 * (-1)^2 + 7 * (-1) + 13 =>
-1 - 5 - 7 + 13 =>
-13 + 13 =>
0
So, (x + 1) is a factor
(x^3 - 5x^2 + 7x + 13) / (x + 1)
Now, when I divide on here, I don't do the whole synthetic division thing that they teach you, namely because Yahoo!Answers formats in a funny way that I'm not too comfortable with. So, what I do is follow these steps:
1) Divide the highest powered term in the numerator by the highest powered term in the denominator
2) Multiply the number I got in step 1 with the denominator
3) Subtract that from the numerator
4) Repeat steps 1 through 3, until I can't do it anymore
x^3 / x = x^2
x^2 * (x + 1) = x^3 + x^2
x^3 - 5x^2 - (x^3 + x^2) => x^3 - 5x^2 - x^3 - x^2 = -6x^2
-6x^2 / x = -6x
-6x * (x + 1) = -6x^2 - 6x
-6x^2 + 7x - (-6x^2 - 6x) => -6x^2 + 7x + 6x + 6x = 13x
13x / x = 13
13 * (x + 1) = 13x + 13
13x + 13 - (13x + 13) = 13x + 13 - 13x - 13 = 0
(x + 1) * (x^2 - 6x + 13) = (x^3 - 5x^2 + 7x + 13)
Now we can use the quadratic formula to find the other values for x
x^2 - 6x + 13 = 0
x = (6 +/- sqrt(36 - 52)) / 2
x = (6 +/- sqrt(-16)) / 2
x = (6 +/- 4i) / 2
x = 3 +/- 2i
x = -1 , 3 + 2i , 3 - 2i