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3 years ago

A polynomial has a leading coefficient of 1 and the

Mathematics

2 answers:

Hoochie [10]3 years ago

7 0

Answer:

(A) $[x-(2+i)][x-(2-i)][x-\sqrt{2}][x+\sqrt{2}]$

Step-by-step explanation:

A polynomial has a leading coefficient of 1 and the following factors with multiplicity 1:

$x-(2+i)\\x-\sqrt{2}$

We apply the following to find the factored form of the polynomial.

If a complex number is a root of a polynomial with real coefficients, its complex conjugate is also a root of that polynomial.

If the polynomial has an irrational root $a+\sqrt{b}$ , where a and b are rational and b is not a perfect square, then it has also a conjugate root $a-\sqrt{b}$ .

$\text{Complex conjugate of }x-(2+i)=x-(2-i)\\\\\text{Complex conjugate of }x-\sqrt{2}=x+\sqrt{2}$

Therefore, the factored form of the polynomial is:

$[x-(2+i)][x-(2-i)][x-\sqrt{2}][x+\sqrt{2}]$

Nutka1998 [239]3 years ago

7 0

Answer:C

Step-by-step explanation:

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