Obtain all zeros of f(x)= x3 + 13x2 + 32x + 20, if one of it's zero is -2
Thnxx for Posting this was very much useful
If alpha and beta are zeros of the quadratic polynomial x square - 6 X + a, find the value of a if 3 alpha + 2 equal to 20
To find all the zeros of a polynomial, we can use the Remainder Theorem and synthetic division.
Given that one of the zeros is -2, we can set up the synthetic division as follows:
-2 | 1 13 32 20
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We'll start by dividing the first coefficient, which is 1, by -2, the root we're testing:
-2 | 1 13 32 20
| -2
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1
Next, we multiply the -2 by the quotient we obtained (which is 1) and add it to the next coefficient (13):
-2 | 1 13 32 20
| -2
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1 11
We repeat this process with the new result, multiplying it by -2 (-2 * 11 = -22), and adding it to the next coefficient (32):
-2 | 1 13 32 20
| -2 22
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1 11 10
Finally, we multiply the -2 by the new result we obtained (which is 10) and add it to the last coefficient (20):
-2 | 1 13 32 20
| -2 22 -20
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1 11 10 0
The remainder obtained is 0, which means that -2 is indeed a zero of the polynomial f(x).
Now, we can write the quotient of the synthetic division: 1x^2 + 11x + 10. This polynomial is the result of dividing f(x) by (x + 2).
To find the remaining zeros, we solve the quadratic equation 1x^2 + 11x + 10 = 0. Factoring this quadratic equation or using the quadratic formula will give us the remaining zeros.
Factoring the quadratic equation: (x + 1)(x + 10) = 0
Setting each factor equal to zero and solving for x:
x + 1 = 0 --> x = -1
x + 10 = 0 --> x = -10
Therefore, the zeros of the polynomial f(x) = x^3 + 13x^2 + 32x + 20 are -2, -1, and -10.
This was very much helpful. Thanxxxxsssss
using either long division by (x+2) or by synthetic division it is easy to show that
f(x)= x3 + 13x2 + 32x + 20
is
f(x) = (x+2)(x^2 + 11x + 10) , then
f(x) = (x+2)(x+10)(x+1)
so the zeros are -2, -10 and -1