## anonymous one year ago 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?

1. campbell_st

so the polynomial has 4 roots.. the real roots are x = 4 and x = -2 the complex roots are $x = -1 \pm i$

2. 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

3. anonymous

Okay.

4. anonymous

2i, 5, -5?

5. campbell_st

is that another question...?

6. anonymous

No, is that the correct answers @campbell_st

7. 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

8. anonymous

The answer I come up with is not one of the options. X^4-10x^2-20x-16

9. Mertsj

f(x)=(x-4)(x+2)[x-(-1+2i)][x-(-1-2i)]

10. Mertsj

11. anonymous

Okay I got x^4+6.5x^2-26x

12. anonymous

Is that correct?

13. 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

14. Mertsj

That's what I get from multiplying those 4 factors. Is that one of your choices?

15. anonymous

Yes, I must have done it wrong. Thank you.

16. anonymous

Wait, no.

17. anonymous

what the hell

18. 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

19. 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

20. 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

21. Mertsj

Is that one of your choices?

22. anonymous

No.

23. anonymous

@Mertsj

24. Mertsj

25. 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

26. 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

27. Mertsj

Before I used x^2-2x+8 instead of x^2-2x-8

28. anonymous

Ah I see.

29. anonymous

Thanks.

30. Mertsj

yw

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