Question

Find a polynomial function with real coefficients of degree 4 And zeros x=1-i and x=√(7). Let the leading coefficient be equal to 1. Written in factored form over rational numbers.

Answer 1

Answer:

$p(x)=(x^{2} -2x+2)(x^{2} -7)$

Step-by-step explanation:

Given that one of the zeroes is 1 - i.

Complex roots occur in pairs.

This means, if a + bi is a zero of a polynomial, then a - bi is also its zero.

Therefore, 1 + i is also a zero of the polynomial.

Factors are x - (1 - i) and x - (1 + i).

Multiply the factors.

[x - (1 - i)][x - (1 + i)] = (x - 1 + i)(x - 1 - i)

$=(x-1)^{2} -i^{2}$

$=(x-1)^{2} +1$

$=x^{2} -2x+1+1$

$=x^{2} -2x+2$

Given that one of the zeroes is $\sqrt{7}$.

Irrational roots occur in pairs.

This means, if an irrational number m is a zero of a polynomial, then - m is also its zero.

Therefore, $-\sqrt{7}$ is also a zero of the polynomial.

Factors are $x-\sqrt{7}$ and $x+\sqrt{7}$.

Multiply the factors.

$(x-\sqrt{7} )(x+\sqrt{7} )$

$=x^{2} -\sqrt{7} ^{2}$

$=x^{2} -7$

Hence, the required polynomial is $p(x)=(x^{2} -2x+2)(x^{2} -7)$.