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Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 5, -3, -1+3i

Respuesta :

mreijepatmat
This is a polynomial of degree 4, because you have 4 roots real or imaginary:1st) x= 5 →(x-5)=0
2nd) x = - 3→(x+3) =0
3rd) x = -1+3i →(x+1-3i) = 0
and the 4th one that is not mentioned which is the conjugate of 
-1+3i, that is -1-3i (in any polynomial if a root has the form of a+bi, there is always a conjugate root = a-bi)
 
4th) x= -1-3i→(x+1+3i) = 0
Hence the polynomial =(x-5)(x+3)((x+1-3i)(x+1+3i)
Solving the above will give you (unless I am mistaken, pls recalculate):

x⁴-9x²-50x-150

A polynomial function of minimum degree with real coefficients whose zeros are 5,-3,-1+3i is

[tex]x^4-9x^2-50x-150[/tex]

Given :

The zeros of the polynomial are [tex]5,-3,-1+3i[/tex]

Complex zeros occurs in pairs. [tex]-1+3i[/tex] is one of the zero

then other zero is [tex]-1-3i[/tex]

So there are 4 zeros for the polynomial

if 'a' is the zero then (x-a) is a factor

Lets write all the zeros in factor

[tex](x-5)(x-(-3))(x-(-1+3i))(x-(-1-3i))\\\left(x-5\right)\left(x+3\right)\left(x+1-3i\right)\left(x+1+3i\right)\\[/tex]

Now we multiply first two parenthesis

[tex]\left(x^2-2x-15\right)\left(x+1-3i\right)\left(x+1+3i\right)[/tex]

Now we multiply last two parenthesis

we apply

[tex]\left(x+1-3i\right)\left(x+1+3i\right)\\a^2-b^2=(a+b)(a-b)\\\left(x+1\right)^2+\left(-3\right)^2\\x^2+2x+1+9\\x^2+2x+10[/tex]

Now we multiply

[tex](x^2-2x-10)(x^2+2x+10)\\x^4-9x^2-50x-150[/tex]

a polynomial function of minimum degree with real coefficients whose zeros are 5,-3,-1+3i is

[tex]x^4-9x^2-50x-150[/tex]

Learn more :  brainly.com/question/7619478

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