Question

Show by factor theorem that \[2{x^4} - 6{x^3} + 3{x^2} + 3x - 2\] is exactly divisible by ${x^2} - 3x + 2$.

Answer 1

Hint: First of all, we will write the factor theorem. Then put in ${x^2} - 3x + 2 = 0$ and see the both roots of this function and fill in those roots one by one in the function \[2{x^4} - 6{x^3} + 3{x^2} + 3x - 2\] and they must be equal to 0.

Complete step-by-step answer:
Let us first look at what factor theorem says:
The factor theorem states that a polynomial $f(x)$ has a factor $(x - k)$ if and only if $f(k) = 0$ (i.e. k is a root)
Now, let \[f(x) = 2{x^4} - 6{x^3} + 3{x^2} + 3x - 2\] and $g(x) = {x^2} - 3x + 2$.
Now, applying the factor theorem, putting $g(x) = 0$.
$ \Rightarrow {x^2} - 3x + 2 = 0$
Rewriting this as:-
$ \Rightarrow {x^2} - x - 2x + 2 = 0$
Rewriting it again as follows:-
$ \Rightarrow x(x - 1) - 2(x - 1) = 0$
Taking the factor common from this, we will get:-
$ \Rightarrow (x - 2)(x - 1) = 0$
Therefore, we have $x = 1$ and $x = 2$.
Now, we will put these one by one in $f(x)$.
Putting $x = 1$ in \[f(x) = 2{x^4} - 6{x^3} + 3{x^2} + 3x - 2\]
So, $f(1) = 2 - 6 + 3 + 3 - 2 = 0$
Now, let us see for $x = 2$.
Putting $x = 2$ in \[f(x) = 2{x^4} - 6{x^3} + 3{x^2} + 3x - 2\]
So, $f(2) = 2{(2)^4} - 6{(2)^3} + 3{(2)^2} + 3 \times 2 - 2$
Simplifying it to get as follows:-
$ \Rightarrow f(2) = 32 - 48 + 12 + 6 - 2$
Simplifying the calculations on RHS, we will get:-
$ \Rightarrow f(2) = 0$
Hence, $(x - 1)(x - 2)$ is a factor of \[2{x^4} - 6{x^3} + 3{x^2} + 3x - 2\].
So, ${x^2} - 3x + 2$ is a factor of \[2{x^4} - 6{x^3} + 3{x^2} + 3x - 2\].

Hence, \[2{x^4} - 6{x^3} + 3{x^2} + 3x - 2\] is exactly divisible by ${x^2} - 3x + 2$.

Note: The students might make the mistake of directly trying to divide the polynomial by the given one but you must take care that you need to use factor theorem.
In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is a special case of the polynomial remainder theorem.
The Remainder Theorem: When we divide a polynomial $f(x)$ by $x - c$ the remainder is $f(c)$.
The point of the Factor Theorem is the reverse of the Remainder Theorem: If you synthetic-divide a polynomial by $x = a$ and get a zero remainder, then, not only is $x = a$ zero of the polynomial (courtesy of the Remainder Theorem), but $x - a$ is also a factor of the polynomial (courtesy of the Factor Theorem).
The remainder theorem and the factor theorem just makes our life super easy by cutting down the number of calculations we need to do. Instead of actually dividing the polynomials, we can directly use them to get our answer.