Question

By actual division, show that ${x^2} - 3$ is a factor of \[2{x^4} + 3{x^3} - 2{x^2} - 9x - 12\].

Answer 1

Hint: In the given question, we are required to divide the polynomial $p\left( x \right) = 2{x^4} + 3{x^3} - 2{x^2} - 9x - 12$ by another polynomial $g\left( x \right) = {x^2} - 3$ to find the quotient and remainder in the division process and find out if one polynomial is a factor of another. So, we will use the long division algorithm for division of $p\left( x \right) = 2{x^4} + 3{x^3} - 2{x^2} - 9x - 12$ by $g\left( x \right) = {x^2} - 3$ where $p\left( x \right)$ is dividend and $g\left( x \right)$ is divisor and we will obtain $q\left( x \right)$ as quotient and $r\left( x \right)$ as remainder. If the remainder comes out to be zero, then the polynomial $g\left( x \right)$ is a divisor of $p\left( x \right)$.

Complete step by step answer:
in the problem, dividend $ = p\left( x \right) = 2{x^4} + 3{x^3} - 2{x^2} - 9x - 12$
Divisor $ = g\left( x \right) = {x^2} - 3$
Now, we will follow the long division method to divide one polynomial by another and to find the quotient and remainder.
\[{x^2} - 3\overset{2{x}^{2}+3x+4}{\overline{\left){\begin{align}
  & {2{x}^{4}}+3{{x}^{3}}-2{{x}^{2}}-9x-12 \\
 & \underline{2{{x}^{4}}-6{x^{2}}} \\
 & 3{{x}^{3}}+4{x^{2}}-9x-12 \\
 & \underline{3{{x}^{3}}-9x} \\
 & 4{x^{2}}-12\\
& \underline{4{{x}^{2}}-12} \\
&0
\end{align}}\right.}}\]
Hence, we have divided the polynomial $p\left( x \right) = 2{x^4} + 3{x^3} - 2{x^2} - 9x - 12$ by $g\left( x \right) = {x^2} - 3$ using the long division method and obtained the quotient polynomial as $q\left( x \right) = 2{x^2} + 3x + 4$ and remainder polynomial as $r\left( x \right) = 0$.

Hence, the quotient is equal to $q\left( x \right) = 2{x^2} + 3x + 4$ and the remaining polynomial is equal to $r\left( x \right) = 0$. Now, since the remainder is equal to zero, we conclude that ${x^2} - 3$ is a factor of $2{x^4} + 3{x^3} - 2{x^2} - 9x - 12$.

Note: The degree of dividend given to us is $4$ and the degree of the divisor is $2$.So, we must remember that the degree of the remainder must be less than that of the divisor. So, in this case, the degree of remainder obtained is $0$ as zero is a constant polynomial. The property of long division \[Dividend = Divisor \times Quotient + Remainder\] holds true for division of numbers as well as polynomials. So, we can also verify the long division as $2{x^4} + 3{x^3} - 2{x^2} - 9x - 12 = \left( {{x^2} - 3} \right) \times \left( {2{x^2} + 3x + 4} \right) + 0$
Multiplying both the brackets on right side of equation, we get,
$ \Rightarrow 2{x^4} + 3{x^3} - 2{x^2} - 9x - 12 = {x^2}\left( {2{x^2} + 3x + 4} \right) - 3\left( {2{x^2} + 3x + 4} \right)$
$ \Rightarrow 2{x^4} + 3{x^3} - 2{x^2} - 9x - 12 = 2{x^4} + 3{x^3} + 4{x^2} - 6{x^2} - 9x - 12$
Adding up like terms on right side of equation, we get,
$ \Rightarrow 2{x^4} + 3{x^3} - 2{x^2} - 9x - 12 = 2{x^4} + 3{x^3} - 2{x^2} - 9x - 12$
Hence, verified.