According to the fundamental theorem of algebra, how many roots does the polynomial \[f\left( x \right)={{x}^{4}}+3{{x}^{2}}+7\] have over the complex numbers, and counting roots with multiplicity greater than one as distinct? (i.e. $f\left( x \right)={{x}^{2}}$ has two roots, both are zero).

Question

According to the fundamental theorem of algebra, how many roots does the polynomial \[f\left( x \right)={{x}^{4}}+3{{x}^{2}}+7\] have over the complex numbers, and counting roots with multiplicity greater than one as distinct? (i.e. $f\left( x \right)={{x}^{2}}$ has two roots, both are zero).

Vedantu Content Team · Accepted Answer

Hint: We start solving the problem by recalling the fundamental theorem of algebra as that every non constant single variable polynomial of degree $n>0$ with complex coefficients has exactly n complex roots counting multiplicity. We then determine whether the given polynomial satisfies the properties of degree and coefficients required for using fundamental theorem of algebra to find the required answer.Complete step by step answer:According to the problem, we are asked to find the number of complex roots that the polynomial $$f\left( x \right)={{x}^{4}}+3{{x}^{2}}+7$$ does have.Let us recall the fundamental theorem of algebra. We know that the fundamental theorem of algebra states that every non constant single variable polynomial of degree $n>0$ with complex coefficients has exactly n complex roots counting multiplicity.We can see that the polynomial is of degree 4 which is greater than 0 and the coefficients are complex (as the real numbers are also part of complex numbers). So, we will have four complex zeros which means the multiplicity is 4.$\therefore $ The polynomial $$f\left( x \right)={{x}^{4}}+3{{x}^{2}}+7$$ have four complex roots.Note: We can also solve this problem as shown below:We have given the polynomial $$f\left( x \right)={{x}^{4}}+3{{x}^{2}}+7$$.Let us assume $y={{x}^{2}}$ and find the zeroes of the resulting polynomial. So, we get $${{y}^{2}}+3y+7=0$$.Let us find the discriminant of $${{y}^{2}}+3y+7=0$$.So, the discriminant is $\Delta ={{3}^{2}}-4\left( 1 \right)\left( 7 \right)$.$\Rightarrow \Delta =9-28$.$\Rightarrow \Delta =-19<0$, which means that the polynomial $${{y}^{2}}+3y+7=0$$ has two complex numbers.We know the square root of a complex number gives two different complex numbers. Which means that we get four complex roots for the polynomial $$f\left( x \right)={{x}^{4}}+3{{x}^{2}}+7$$.So, the multiplicity of the roots of the polynomial $$f\left( x \right)={{x}^{4}}+3{{x}^{2}}+7$$ is 4.