anonymous
  • anonymous
Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 4, -2, and -1 + 2i How would I solve this?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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campbell_st
  • campbell_st
so the polynomial has 4 roots.. the real roots are x = 4 and x = -2 the complex roots are \[x = -1 \pm i\]
campbell_st
  • campbell_st
so to find the complex roots start with \[x = -1 \pm i\] add 1 to both sides \[x + 1 = \pm i\] square both sides of the equation \[(x + 1)^2 = i^2\] remember i^2 = -1 then \[x^2 + 2x + 1 = -1\] add 1 to both sides of the equation and you will have the quadratic factor of the polynomial... so the polynomial is \[P(x) = (x^2 + 2x +1)(x -4)(x+1)\] just distribute for the equation
anonymous
  • anonymous
Okay.

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anonymous
  • anonymous
2i, 5, -5?
campbell_st
  • campbell_st
is that another question...?
anonymous
  • anonymous
No, is that the correct answers @campbell_st
campbell_st
  • campbell_st
well the polynomial is \[P(x) = (x^2 + 2x + 2)(x -4)(x +2)\] which becomes \[P(x) = (x^2+ 2x +2)(x^2 -2x - 8)\] you just need to multiply the 2 quadratic factors together
anonymous
  • anonymous
The answer I come up with is not one of the options. X^4-10x^2-20x-16
Mertsj
  • Mertsj
f(x)=(x-4)(x+2)[x-(-1+2i)][x-(-1-2i)]
Mertsj
  • Mertsj
Multiply the four factors and that will be your answer.
anonymous
  • anonymous
Okay I got x^4+6.5x^2-26x
anonymous
  • anonymous
Is that correct?
Mertsj
  • Mertsj
f(x)=(x^2-2x-8)(x^2+5)=x^4-2x^3-8x^2+5x^2-10x-40=x^4-2x^3-3x^2-10x-40
Mertsj
  • Mertsj
That's what I get from multiplying those 4 factors. Is that one of your choices?
anonymous
  • anonymous
Yes, I must have done it wrong. Thank you.
anonymous
  • anonymous
Wait, no.
anonymous
  • anonymous
what the hell
Mertsj
  • Mertsj
If 4is a root, then x-4 is a factor. If -2 is a root then x-(-2)=x+2 is a factor If -1+2i is a root then -1 - 2i is also a root If -1+2i is a root then x-(-1+2i) is a factor If -1-2i is a root then x - (-1-2i) is a factor
Mertsj
  • Mertsj
So now we have 4 factors of the polynomial: (x-4)(x+2)[x-(-1+2i)][x-(-1-2i)] (x-4)(x+2)=x^2-2x-8 [x-(-1+2i)][x-(-1-2i)]=x^2-x(-1-2i)-x(-1+2i)+(-1+2i)(-1-2i)= x^2+x+2xi+x-2xi+5=x^2+2x+5
Mertsj
  • Mertsj
So now we must multiply (x^2-2x+8)(x^2+2x+5)= x^4+2x^3+5x^2-2x^3-4x^2-10x+8x^2+16x+40= x^4+9x^2+6x+40
Mertsj
  • Mertsj
Is that one of your choices?
anonymous
  • anonymous
No.
anonymous
  • anonymous
@Mertsj
Mertsj
  • Mertsj
What are the answer choices?
anonymous
  • anonymous
a. f(x) = x4 - 7x2 - 26x - 40 b. f(x) = x4 - 3x3 - 8x2 - 13x - 40 c. f(x) = x4 + 6.5x2 - 26x - 40 d. f(x) = x4 - 3x3 + 8x2 + 13x + 40
Mertsj
  • Mertsj
I made a typo when I multiplied: (x^2-2x-8)(x^2+2x+5)=x^4+2x^3+5x^2-2x^3-4x^2-10x-8x^2-16x-40= x^4-7x^2-26x-40
Mertsj
  • Mertsj
Before I used x^2-2x+8 instead of x^2-2x-8
anonymous
  • anonymous
Ah I see.
anonymous
  • anonymous
Thanks.
Mertsj
  • Mertsj
yw

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