SOLUTION: Find a polynomial function of least degree having only real​ coefficients, a leading coefficient of​ 1, and zeros of 3 and 2 plus i .

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Question 1126012: Find a polynomial function of least degree having only real​ coefficients, a leading coefficient of​ 1, and zeros of 3 and 2 plus i .

Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

if zeros are x%5B1%5D=3 (real one) and x%5B2%5D=2+%2B+i (complex one), then you also have x%5B3%5D=2+-+i because complex zeros always come in pairs
since you have three zeros, a polynomial function of least degree will be a polynomial function of degree 3
f%28x%29=%28x-x%5B1%5D%29%28x-x%5B2%5D%29%28x-x%5B3%5D%29...substitute given zeros
f%28x%29=%28x-3%29%28x-%282+%2B+i%29%29%28x-%282+-+i%29%29
f%28x%29=%28x-3%29%28x-2+-i%29%28x-2+%2B+i%29
f%28x%29=%28x-3%29%28x%5E2+-+4+x+%2B+5%29
f%28x%29=x%5E3+-+7+x%5E2+%2B+17+x+-+15

Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


Given the three roots p, q, and r, one way you can find the equation is multiply out (x-p)(x-q)(x-r), as shown by the other tutor. With the two complex roots, the algebra gets a bit ugly.

So here is another way to find the equation that some students might find easier.

In the final equation (with leading coefficient 1)

f%28x%29+=+x%5E3%2Bbx%5E2%2Bcx%2Bd,

the coefficient of the quadratic term is the opposite of the sum of the roots:
b+=+-%28p%2Bq%2Br%29

the coefficient of the linear term is the sum of the products of the roots two at a time:
c+=+pq%2Bpr%2Bqr

and the constant term is the opposite of the product of the three roots:
d+=+-pqr

With roots 3, 2+i, and 2-i, we have...

p%2Bq%2Br+=+3%2B%282%2Bi%29%2B%282-i%29+=+7 so b = -7;

pq%2Bpr%2Bqr+=+%286%2B3i%29%2B%286-3i%29%2B5+=+17 so c = 17; and

pqr+=+3%285%29+=+15 so d = -15.

Then the polynomial is

f%28x%29+=+x%5E3-7x%5E2%2B17x-15