Find a degree 3 polynomial with real coefficients having zeros 1 and 4 i and a lead coefficient of 1. Write P in expanded form

Answer 1

Since this should be a "degree 3" polynomial, then we know that the highest power of x must be 3. So our polynomial should look something like this:

P = a(x – c₁)(x – c₂)(x – c₃),

where a is the leading coefficient, and c₁, c₂, and c₃ are the zeros (also known as the x-intercepts) of the polynomial.

According to the instructions, the leading coefficient of this polynomial should be 1, so we will set a = 1.

As for the zeros, we were given two of them... 1 and 4i. Since 4i is a complex number (because of the i in the number), then we can apply the theorem that says that if a complex number is a zero of a polynomial function, then the conjugate of that complex number is also a zero. A complex number is usually written in the form a + bi (where a and b are real numbers). The conjugate of the complex number a + bi is just a – bi. Notice that only the sign in the middle changed.

So, the zero that we already know, 4i, could be written as 0 + 4i. Thus, the conjugate of 0 + 4i must be 0 – 4i (which, of course, is just –4i). This conjugate is also a zero of our polynomial. So now we know that the three zeros of our function must be:

c₁ = 1

c₂ = 4i

c₃ = –4i

Now let's substitute these values (along with our leading coefficient a = 1) into the polynomial formula from before.

P = 1(x – 1)(x – 4i)(x + 4i)

Finally, we need to write this in expanded form, with only real number coefficients (no complex numbers). These three factors can be multiplied in any order, but I would suggest beginning by multiplying (x – 4i)(x + 4i), using the FOIL method:

(x – 4i)(x + 4i)

= x² + 4i x – 4i x – 16i²

= x² – 16(-1) (because, remember, i² = -1)

= x² + 16

So now, our polynomial looks like this:

P = (x – 1)(x² + 16)

Notice that I don't have to keep writing the factor of 1 in front. If that was any number other than 1, I would need to keep it in the front, and multiply it later.

So let's finish this off by using the FOIL method once again.

P = (x – 1)(x² + 16)

= x³ + 16x – x² – 16

= x³ – x² + 16x – 16. (Final answer)

Answer 2

one factor is x-1

4i and -4i are the two other zeros. Imaginary zeros come in conjugate pairs

x=4i

x^2 = -16

x^2 +16 is the other factor

multiply them: (x-1)(x^2+16) = P(x) = x^3 -x^2 +16x -16