Question

Can ${{x}^{2}}-1$ be the quotient on division of ${{x}^{6}}+2{{x}^{3}}+x-1$ by a polynomial in $'x'$ of degree 5.

Answer 1

We solve this problem by using the long division method considering the divisor as a general polynomial of degree 5. The general representation of polynomial of degree 5 is
$g\left( x \right)=a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}+d{{x}^{2}}+ex+f$ where $a,b,c,d,e,f$ are real numbers and $a\ne 0$
We divide the given polynomial with this polynomial and check the quotient whether it is possible to get the given quotient.
We use the definition of a division that is
$\text{Dividend}=\left( \text{Divisor} \right)\times \left( \text{Quotient} \right)+\left( \text{Remainder} \right)$

Complete answer:
We are given that the polynomial as ${{x}^{6}}+2{{x}^{3}}+x-1$
Let us assume that the given polynomial as
$\Rightarrow f\left( x \right)={{x}^{6}}+2{{x}^{3}}+x-1$
We are given that this polynomial is divided by a polynomial of degree 5
We know that the general representation of a polynomial of degree 5 as
$g\left( x \right)=a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}+d{{x}^{2}}+ex+f$ where $a,b,c,d,e,f$ are real numbers and $a\ne 0$
Let us assume that the quotient obtained after dividing the given polynomial with this polynomial as $q\left( x \right)$ and the remainder as $r\left( x \right)$
We know that the definition of a division that is
$\text{Dividend}=\left( \text{Divisor} \right)\times \left( \text{Quotient} \right)+\left( \text{Remainder} \right)$
Now, by using the above definition to given polynomial we get

& \Rightarrow f\left( x \right)=g\left( x \right)\times q\left( x \right)+r\left( x \right) \\\ & \Rightarrow {{x}^{6}}+2{{x}^{3}}+x-1=\left( a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}+d{{x}^{2}}+ex+f \right)\times q\left( x \right)+r\left( x \right) \\\ \end{aligned}$$ Now, we use the long division method. We know that in long division methods we eliminate the terms of dividend from higher degree to lower degree. Now, let us use the long division method that is by eliminating the first term in the dividend that is $${{x}^{6}}$$ by multiplying the divisor with $$\dfrac{x}{a}$$ we get the quotient and the remainder as $$\left( a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}+d{{x}^{2}}+ex+f \right)\overset{\dfrac{x}{a}}{\overline{\left){\begin{aligned} & \left( {{x}^{6}}+2{{x}^{3}}+x-1 \right) \\\ & -\left( {{x}^{6}}+\dfrac{b}{a}{{x}^{5}}+\dfrac{c}{a}{{x}^{4}}+\dfrac{d}{a}{{x}^{3}}+\dfrac{e}{a}{{x}^{2}}+\dfrac{f}{a}x \right) \\\ & =\left( \dfrac{-b{{x}^{5}}}{a}-\dfrac{c{{x}^{4}}}{a}+{{x}^{3}}\left( 2-\dfrac{d}{a} \right)-\dfrac{e{{x}^{2}}}{a}+x\left( 1-\dfrac{f}{a} \right)-1 \right) \\\ \end{aligned}}\right.}}$$ By using the division algorithm that is $$Dividend=divisor\times quotient+remainder$$ to above division then we get $$\Rightarrow \left( {{x}^{6}}+2{{x}^{3}}+x-1 \right)=\left( a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}+d{{x}^{2}}+ex+f \right)\times \left( \dfrac{x}{a} \right)+\left( \dfrac{-b{{x}^{5}}}{a}-\dfrac{c{{x}^{4}}}{a}+{{x}^{3}}\left( 2-\dfrac{d}{a} \right)-\dfrac{e{{x}^{2}}}{a}+x\left( 1-\dfrac{f}{a} \right)-1 \right)$$ Here, we can see that there is the remainder in the above division which can be further divided by divisor. Now, by eliminating the first term in the remainder in above equation that is by multiplying the divisor with $$\dfrac{-{{b}^{2}}}{{{a}^{2}}}$$ we get the quotient and remainder as $$\left( a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}+d{{x}^{2}}+ex+f \right)\overset{\dfrac{x}{a}-\dfrac{{{b}^{2}}}{{{a}^{2}}}}{\overline{\left){\begin{aligned} & \left( {{x}^{6}}+2{{x}^{3}}+x-1 \right) \\\ & -\left( {{x}^{6}}+\dfrac{b}{a}{{x}^{5}}+\dfrac{c}{a}{{x}^{4}}+\dfrac{d}{a}{{x}^{3}}+\dfrac{e}{a}{{x}^{2}}+\dfrac{f}{a}x \right) \\\ & =\left( \dfrac{-b{{x}^{5}}}{a}-\dfrac{c{{x}^{4}}}{a}+{{x}^{3}}\left( 2-\dfrac{d}{a} \right)-\dfrac{e{{x}^{2}}}{a}+x\left( 1-\dfrac{f}{a} \right)-1 \right) \\\ & -\left( -\dfrac{b}{a}{{x}^{5}}-\dfrac{{{b}^{2}}}{{{a}^{2}}}{{x}^{4}}-\dfrac{cb}{{{a}^{2}}}{{x}^{3}}-\dfrac{db}{{{a}^{2}}}{{x}^{2}}-\dfrac{ed}{{{a}^{2}}}x-\dfrac{fb}{{{a}^{2}}} \right) \\\ & =\left( {{x}^{4}}\left( \dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{c}{a} \right)+k \right) \\\ \end{aligned}}\right.}}$$ By using the division algorithm that is $$Dividend=divisor\times quotient+remainder$$ to above division then we get $$\Rightarrow \left( {{x}^{6}}+2{{x}^{3}}+x-1 \right)=\left( a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}+d{{x}^{2}}+ex+f \right)\times \left( \dfrac{x}{a}-\dfrac{b}{{{a}^{2}}} \right)+\left( {{x}^{4}}\left( \dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{c}{a} \right)+k \right)$$ Here, we can see $$'k'$$ is the remaining remainder which cannot be useful. Here we can say that the quotient is given as $$\Rightarrow q\left( x \right)=\left( \dfrac{x}{a}-\dfrac{b}{{{a}^{2}}} \right)$$ Here, we can see that the quotient has a degree of 4 and the divisor has the degree of 5. So, we can stop the division here. Here, we can see that the quotient as $$\left( \dfrac{x}{a}-\dfrac{b}{{{a}^{2}}} \right)$$ We are asked to find whether $${{x}^{2}}-1$$ can be quotient which is a polynomial of degree 2 but we get the polynomial of degree 1 as quotient in the division. As we can see that linear polynomial and the polynomial of degree 2 can never be equal we can say that $${{x}^{2}}-1$$ will not be quotient when $${{x}^{6}}+2{{x}^{3}}+x-1$$ by a polynomial in $$'x'$$ of degree 5 Therefore, the answer is ‘not possible’. **Note:** We have another explanation by using the definition of polynomials. When a polynomial of degree $$'n'$$ is divided by a polynomial of degree $$'r'$$ then the quotient has the degree of $$'n-r'$$ Here we are given that the polynomial of degree 6 is divided by polynomial of degree 5. So, we can say that the degree of quotient is $$6-5=1$$ But we are asked to check the polynomial $${{x}^{2}}-1$$ which has degree 2, but we get the degree of quotient as 1 which never equals each other. So, we can say that $${{x}^{2}}-1$$ will not be quotient when $${{x}^{6}}+2{{x}^{3}}+x-1$$ by a polynomial in $$'x'$$ of degree 5.