Forming a polynomial f(x) with real coefficients having the given degree and zeros

[Steph Annie]

Forming a polynomial f(x) with real coefficients having the given degree 4, zeros 5-3i; -3 and multiplicity 2.
In addition to the zeros of f(x) already given, -2+5i is also a zero.

Using the FACTOR THEOREM.
f(x)= a (x-r₁₎(x-r₂)(x-r₃)(x-r₄)

f(x) = a [x-(-3)] [x-(-3)][x-(5-3i)][x-(5+3i)]

f(x) = a(x²+6x+9)[x²-(5-3i)x - (5 + 3i)x + (5-3i)]

f(x)=a(x² +6x +9) [x²-5x +3ix -5x -3ix +25 + 25i - 15i -9i^2]
=a(x²+6x+9)(x²-10x + 34)
=a (x⁴-10x³+34x²+6x³-60x²+204x+9x²-90x+306
= a(x⁴-4x³-17x²+114x+306)

Final answer: The fourth degree polynomial with zeros 5-3i, 5+3i, -3 and multiplicity 2 is f(x)= a(x⁴-4x³-17x²+114x+306).

I am having trouble seeing how they did the factoring. This is an example problem, and I need to understand the form they are using to do my homework problems. For starters, how do they get x²+6x+9 ? Could you show me those steps? What exactly are they plugging into a (x-r₁₎(x-r₂)(x-r₃)(x-r₄)??

I'm at a loss here and all these coefficients and degrees look very confusing. If you could break it down a little for me or show me a simpler way to form the polynomial using the factor theorem I'd really appreciate it.

 

[Deleted member 4993]

Forming a polynomial f(x) with real coefficients having the given degree 4, zeros 5-3i; -3 and multiplicity 2.
In addition to the zeros of f(x) already given, -2+5i is also a zero.

Using the FACTOR THEOREM.
f(x)= a (x-r₁₎(x-r₂)(x-r₃)(x-r₄)

f(x) = a [x-(-3)] [x-(-3)][x-(5-3i)][x-(5+3i)]

f(x) = a(x²+6x+9)[x²-(5-3i)x - (5 + 3i)x + (5-3i)]

f(x)=a(x² +6x +9) [x²-5x +3ix -5x -3ix +25 + 25i - 15i -9i^2]
=a(x²+6x+9)(x²-10x + 34)
=a (x⁴-10x³+34x²+6x³-60x²+204x+9x²-90x+306
= a(x⁴-4x³-17x²+114x+306)

Final answer: The fourth degree polynomial with zeros 5-3i, 5+3i, -3 and multiplicity 2 is f(x)= a(x⁴-4x³-17x²+114x+306).

I am having trouble seeing how they did the factoring. This is an example problem, and I need to understand the form they are using to do my homework problems. For starters, how do they get x²+6x+9 ? Could you show me those steps? What exactly are they plugging into a (x-r₁₎(x-r₂)(x-r₃)(x-r₄)??

I'm at a loss here and all these coefficients and degrees look very confusing. If you could break it down a little for me or show me a simpler way to form the polynomial using the factor theorem I'd really appreciate it.

(a+b)² = a² + 2ab + b²
(x + 3)² = x² + 2*3*x + 3²

 

[Steph Annie]

(a+b)² = a² + 2ab + b²
(x + 3)² = x² + 2*3*x + 3²

ha! Okay. I see. I have this stupid habit of seeing (x+3)(x+3) and thinking of a perfect square, x²- 9, in which case it would be (x+3)(x-3) and that's why I couldn't see where they got x²+ 6x +9. Blah.
Thanks!!

And so, going back to before: a[x-(-3)][x-5+3i][x-5+3i]
[(x² +6x+9)] then [(x²-5x+3ix-25-15i+3xi-15i+9i²)]

All these x's and i's are kinda freaking me out!! Is my distribution correct there?
After I distributed, I combined the like terms from that expression and got x²-5x+6ix-30i+9i²-25.
I can't remember what I should do with the 9i². Do I just leave it there?

 

[Quaid]

Forming a polynomial f(x) with real coefficients having the given degree 4, zeros 5-3i; -3 and multiplicity 2.

In addition to the zeros of f(x) already given, -2+5i is also a zero.

Hi Steph,

-2 + 5i is a zero of f(x)?

I don't understand how a fourth-degree polynomial has six zeros! Is this exercise from a course assignment? If so, what's the course name?

going back to before: a[x-(-3)]^2[x-5+3i][x-5-3i]

[(x² +6x+9)] then [(x²-5x+3ix-25-15i+3xi-15i+9i²)]

I'm guessing that the missing power of two is a typo, but you also have a sign error, when subtracting x-(5+3i).

(Both corrected in red above.)


:idea: Don't use the word "then" to represent multiplication; type an asterisk, instead.


With the sign error fixed, multiply (x-5+3i)*(x-5-3i) again.

The resulting polynomial will have no imaginary parts.

Cheers