
How do you determine whether $x - 1$ is a factor of the polynomial $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$ ?
Answer
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Hint: We have been given a polynomial expression $p\left( x \right)$ and we have to determine whether another expression $x - a$ is a factor of the given polynomial or not. For that, we use remainder theorem. According to that, if $x - a$ is a factor of $p\left( x \right)$ then $p\left( a \right) = 0$ . In this question, first , we equate the $x - a$ to zero and determines the value of $x = a$ . Now we substitute this value of $x$ in the given polynomial $p\left( x \right)$ and check whether the value of the polynomial is $0$ or not. If the obtained value is equal to zero then the expression $x - a$ is the factor of $p\left( x \right)$ otherwise not.
Complete step-by-step solution:
Step1: Given polynomial is $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$ and we have to determine whether $x - 1$ is a factor of the above given polynomial or not. For that, we equate $x - 1 = 0$ so we get the value of $x = 1$
Step2: Now we use remainder theorem to determine whether $x - 1$ is a factor of $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$ or not. For that, we substitute the value of $x = 1$ in the given polynomial, we get
$4{\left( 1 \right)^4} - 2{\left( 1 \right)^3} + 3{\left( 1 \right)^2} - 2\left( 1 \right) + 1$
On simplification we get
$
\Rightarrow 4 \times 1 - 2 \times 1 + 3 \times 1 - 2 \times 1 + 1 \\
\Rightarrow 4 - 2 + 3 - 2 + 1 \\
\Rightarrow 4 \\
$
Step3: Since the obtained value of the given polynomial $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$ at $x = 1$ is not equal to zero.
So $x - 1$ is not the factor of the polynomial $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$.
Note: While evaluating the power of a number remember that the odd power of a negative number results in a negative number, while the even power of a negative number results in a positive number.
In the case of a positive number, the even power or odd power in both cases, results in a positive number.
Complete step-by-step solution:
Step1: Given polynomial is $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$ and we have to determine whether $x - 1$ is a factor of the above given polynomial or not. For that, we equate $x - 1 = 0$ so we get the value of $x = 1$
Step2: Now we use remainder theorem to determine whether $x - 1$ is a factor of $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$ or not. For that, we substitute the value of $x = 1$ in the given polynomial, we get
$4{\left( 1 \right)^4} - 2{\left( 1 \right)^3} + 3{\left( 1 \right)^2} - 2\left( 1 \right) + 1$
On simplification we get
$
\Rightarrow 4 \times 1 - 2 \times 1 + 3 \times 1 - 2 \times 1 + 1 \\
\Rightarrow 4 - 2 + 3 - 2 + 1 \\
\Rightarrow 4 \\
$
Step3: Since the obtained value of the given polynomial $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$ at $x = 1$ is not equal to zero.
So $x - 1$ is not the factor of the polynomial $4{x^4} - 2{x^3} + 3{x^2} - 2x + 1$.
Note: While evaluating the power of a number remember that the odd power of a negative number results in a negative number, while the even power of a negative number results in a positive number.
In the case of a positive number, the even power or odd power in both cases, results in a positive number.
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