Question

How to factorise: $ 2{{h}^{3}}+h-3 $ ?

Answer 1

Hint: We factor the given equation with the help of vanishing method. In this method we find a number $ a $ such that for $ x=a $ , if $ f\left( a \right)=0 $ then $ \left( x-a \right) $ is a root of $ f\left( x \right) $ . We assume $ f\left( h \right)=2{{h}^{3}}+h-3 $ and take the value of $ a $ as 1.

Complete step by step solution:
We find the value of $ h=a $ for which the function $ f\left( h \right)=2{{h}^{3}}+h-3=0 $ .
We take $ h=a=1 $ .
We can see $ f\left( 1 \right)=2\times {{1}^{3}}+1-3=2+1-3=0 $ .
So, the root of the $ f\left( h \right)=2{{h}^{3}}+h-3 $ will be the function $ \left( h-1 \right) $ . This means for $ h=a $ , if $ f\left( a \right)=0 $ then $ \left( h-a \right) $ is a root of $ f\left( h \right) $ .
Therefore, the term $ \left( h-1 \right) $ is a factor of the polynomial $ 2{{h}^{3}}+h-3 $ .
We can now divide the polynomial $ 2{{h}^{3}}+h-3 $ by $ \left( h-1 \right) $ .
\[h-1\overset{2{{h}^{2}}+2h+3}{\overline{\left){\begin{align}
  & 2{{h}^{3}}+h-3 \\
 & \underline{2{{h}^{3}}-2{{h}^{2}}} \\
 & 2{{h}^{2}}+h-3 \\
 & \underline{2{{h}^{2}}-2h} \\
 & 3h-3 \\
 & \underline{3h-3} \\
 & 0 \\
\end{align}}\right.}}\]
We first tried to equate the highest power of the dividend with the highest power of the divisor and that’s why we multiplied with $ 2{{h}^{2}} $ . We get \[2{{h}^{3}}-2{{h}^{2}}\]. We subtract it to get \[2{{h}^{2}}+h-3\]. We again equate with the highest power of the remaining terms. We multiply with $ 2h $ and subtract to get \[3h-3\]. At the end we had to multiply with 3 to complete the division. The quotient is \[2{{h}^{2}}+2h+3\].
The final factorisation is $ 2{{h}^{3}}+h-3=\left( h-1 \right)\left( 2{{h}^{2}}+2h+3 \right) $ .
So, the correct answer is “ $ 2{{h}^{3}}+h-3=\left( h-1 \right)\left( 2{{h}^{2}}+2h+3 \right) $ ”.

Note: We know for a general equation of quadratic $ a{{x}^{2}}+bx+c=0 $ , the value of the roots of x will be $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ .
In the given equation we have \[2{{h}^{2}}+2h+3\]. The values of a, b, c are $ 2,2,3 $ respectively.
We put the values and get x as $ x=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times 2\times 3}}{2\times 2}=\dfrac{-2\pm \sqrt{-20}}{4}=\dfrac{-2\pm 2i\sqrt{5}}{4}=\dfrac{-1\pm i\sqrt{5}}{2} $