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# How do you determine whether $x – 1$ is a factor of the polynomial $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$?

Last updated date: 20th Jun 2024
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Hint: We will use remainder theorem and put x = 1 in the given polynomial. If it comes out to be zero, then $x – 1$ is a factor of the given polynomial, otherwise it is not.

Complete step by step solution:
We are given that we need to determine whether x – 1 is a factor of the polynomial $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$.
Let us call the given polynomial $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$ to be P (x).
So, we have: $P(x) = 4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$.
Now, according to the remainder theorem, if P (1) is 0, then x – 1 is a factor of the given polynomial P (x).
Putting x = 1 in P (x), we will then obtain the following equation:-
$\Rightarrow P(1) = 4{(1)^4} - 2{(1)^3} + 3{(1)^2} - 2(1) + 1$
Simplifying the powers of 1 in the above equation, we will then obtain the following equation:-
$\Rightarrow P(1) = 4(1) - 2(1) + 3(1) - 2(1) + 1$
Simplifying the calculations in the above equation, we will then obtain the following equation:-
$\Rightarrow$P (1) = 4 – 2 + 3 – 2 + 1
Simplifying the calculations further in the above equation, we will then obtain the following equation:-
$\Rightarrow$P (1) = 4

Now, since P (1) is not zero, therefore, x – 1 is not a factor of the given polynomial P(x) which is $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$.

Note: The students must know the definition of the Remainder Theorem which we are discussing above in the question.
Remainder Theorem: According to The Remainder Theorem (x − a) is a factor of a polynomial P (x) if and only if P (a) = 0.
The students must also note that we also may use the actual long division method to find the answer. But this method that is the method of using the “Remainder Theorem” is the easiest because it is just simple calculations.
Let us try that alternate method as well:-
We will actually divide the given polynomial $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$ by (x – 1).
If we multiply (x – 1) by $4{x^3} + 2{x^2} + 5x + 3$, then we will get the remainder of 4 and we cannot divide it any further.
Now, we can clearly see that we get 4 as the remainder, therefore, the given factor (x – 1) is not a factor of P (x).