Use the given information about a polynomial whose coefficients are real numbers to find the remaining zeros.
degree: 6
Zeros: -8 + 11x(can't put this sign in looks like ii with slash over)i, -7 + 17i, 16 - isqrt 2
You know that since the roots come in conjugate pairs, there will be
-7 + 17i and -7 -17i
16 - isqrt2 and 16 + isqrt2
so the factors will be
(x^2 - 14x + 338) that's (x^2 -2*7x + 17^2 + 7^2)
(x^2 - 32x + 258) that's (x^2 - 2*16x + 16^2 + sqrt2^2)
and the other pair you can figure out. I can't tell what you mean.
Hmm. I think that's +14x in the 1st case -(2*-7)
To find the remaining zeros of the polynomial, let's start by using the given information that the coefficients are real numbers. This tells us that complex conjugate pairs of zeros must exist in the polynomial.
First, let's focus on the zero: -8 + 11i.
Since the coefficients are real, the conjugate of -8 + 11i must also be a zero. The conjugate of a complex number is found by changing the sign of the imaginary part. So, the conjugate of -8 + 11i is -8 - 11i.
Therefore, we now have two zeros: -8 + 11i and -8 - 11i.
Moving on to the next zero: -7 + 17i.
Similarly, its conjugate, -7 - 17i, must also be a zero.
So far, we have the following zeros:
-8 + 11i, -8 - 11i, -7 + 17i, -7 - 17i.
Finally, let's consider the zero: 16 - √2.
Since the coefficients are real, we don't need to find its conjugate since it does not involve any imaginary parts.
Therefore, the remaining zeros of the polynomial are:
-8 + 11i, -8 - 11i, -7 + 17i, -7 - 17i, 16 - √2.
These are the six zeros of the polynomial with a degree of 6.