Question

A fourth-degree polynomial with integer coefficients has zeros at 1 and 3 +0" id="TexFormula1" title=" \sqrt{5} " alt=" \sqrt{5} " align="absmiddle" class="latex-formula"> Which number cannot also be a zero of this polynomial?
A. 1
B. -3
C. 3-$\sqrt{5$
D. 3+$\sqrt{2}$

Answer 1

If a radical be a zero of a polynomial then it's conjugate will also be the zero.

Like for the given problem has a zero 3+ √5. So, it's conjugate 3- √5 must be the other zero.

Hence, the zeroes of the polynomial are 1, 3+ √5 and 3-√5.

According to the problem, the degree of the polynomial is 4.

We already have three zeroes, so only one zero is unknown.

Because degree of polynomial = number of zeroes.

Other zeroes can be 1 and -3. We have already found that 3 - √5 is one of the zero.

So, A, B, C cannot be the correct choice.

But if 3+ √2 will be a zero then 3- √2 will also be the zero. Then the polynomial will have 5 zeroes which is not possible as the polynomial has degree four.

So, D: 3+ √2 cannot also be a zero of this polynomial.

Hope this helps you!