Question

Find the H.C.F. of ${{x}^{3}}-7{{x}^{2}}+14x-8,{{x}^{3}}-6{{x}^{2}}+11x-6$ by division method

Answer 1

Hint: H.C.F. of two polynomials can be given by their common factors. Factorize each of the cubic equations given in the problem by guessing one of the roots of them and hence, get the other roots of them by dividing the polynomial by that factor. If $\alpha $ is the root of any polynomial, then $\left( x-\alpha \right)$ is a factor of it.

Step-by-step answer:
As we are given two cubic equations as
${{x}^{3}}-7{{x}^{2}}+14x-8$ ………. (i)
${{x}^{3}}-6{{x}^{2}}+11x-6$………… (ii)
Now, we know H.C.F. of two polynomials means it will be a common factor for both polynomials. So, we need to find factors of both polynomials individually and hence by observing them, we will get H.C.F of them.
So, first cubic is given as
$={{x}^{3}}-7{{x}^{2}}+14x-8$
Put x = 1 to the above equation,
$={{\left( 1 \right)}^{3}}-7{{\left( 1 \right)}^{2}}+14\left( 1 \right)-8$
= 1 – 7 + 14 – 8 = 14 – 14 = 0
Hence x = 1 is the root of the given cubic, as the polynomial becomes 0 at x = 1.
So, x – 1 should be a factor of this polynomial. Hence, we need to divide the given cubic by x -1 to get other factors as
\[x-1\overset{{{x}^{2}}-6x+8}{\overline{\left){\begin{align}
  & {{x}^{3}}-7{{x}^{2}}+14x-8 \\
 & {{x}^{3}}-{{x}^{2}} \\
 & \overline{\begin{align}
  & 6{{x}^{2}}+14x-8 \\
 & \underline{6{{x}^{2}}+6x} \\
\end{align}} \\
 & 8x-8 \\
 & \underline{\left( 8x-8 \right)} \\
 & \text{ 0} \\
\end{align}}\right.}}\]
Hence, the factor of the cubic is ${{x}^{2}}-6x+8$.
So, we can represent the first cubic equation in form of factors only as
$\left( x-1 \right)\left( {{x}^{2}}-6x+8 \right)$
We need to factorize the quadratic ${{x}^{2}}-6x+8$ as well to get the factors in linear form.
So, we can split the middle term of the quadratic to -4, -2. So, we get
$\begin{align}
  & =\left( x-1 \right)\left( {{x}^{2}}-4x-2x+8 \right) \\
 & \left( x-1 \right)\left( x\left( x-4 \right)-2\left( x-4 \right) \right) \\
\end{align}$
$=\left( x-1 \right)\left( x-2 \right)\left( x-4 \right)$…….. (iii).
Now, let us factorize the second cubic. So, we are given second cubic as
${{x}^{3}}-6{{x}^{2}}+11x-6$
Put x = 1, we get
$={{\left( 1 \right)}^{3}}-6{{\left( 1 \right)}^{2}}+11\left( 1 \right)-6$
= 1- 6 + 11 – 6 = 12 – 12
= 0
Hence, x -1 should be a factor of the polynomial, as x = 1 is the root of it.
So, dividing the polynomial by x -1, we get
\[x-1\overset{{{x}^{2}}-5x+6}{\overline{\left){\begin{align}
  & {{x}^{3}}-6{{x}^{2}}+11x-6 \\
 & {{x}^{3}}-{{x}^{2}} \\
 & \overline{\begin{align}
  & 5{{x}^{2}}+11x-6 \\
 & \underline{5{{x}^{2}}+5x} \\
\end{align}} \\
 & 6x-6 \\
 & \underline{6x-6} \\
 & \text{ 0} \\
\end{align}}\right.}}\]
Hence, the cubic can be represented as
$\left( x-1 \right)\left( {{x}^{2}}-5x+6 \right)$
Now, we can split middle-term of the quadratic ${{x}^{2}}-5x+6\to $ -3, -2, to get the other factors of the cubic
=$\begin{align}
  & =\left( x-1 \right)\left( {{x}^{2}}-3x-2x+6 \right) \\
 & =\left( x-1 \right)\left( x\left( x-3 \right)-2\left( x-3 \right) \right) \\
\end{align}$
$=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$…………. (iv)
Now, we can observe the equation (iii) and (iv) and hence, we get that the factors $\left( x-1 \right)\left( x-2 \right)$are common factors of both the cubic. Hence, H.C.F. can be given as
H.C.F = $\left( x-1 \right)\left( x-2 \right)$
H.C.F = ${{x}^{2}}-2x-x+2$
H.C.F = ${{x}^{2}}-3x+2$
So, ${{x}^{2}}-3x+2$ is the correct answer.

Note: One may guess all the roots of the given cubic equations. As the roots are of integer form. But, we need to use a division method to get H.C.F. So, apply the division rule with the question.Don’t confuse the root and factor step. If the root of the cubic is 3 then (x – 3) is the factor of it, don’t take (x + 3) as the factor of cubic, It is the general confusion among students. So, always clear with this concept.