
What is the difference between the remainder theorem and the factor theorem?
Answer
507k+ views
Hint: In the given theorem we just need to answer the difference of the two popular theorems factor theorem and the remainder theorem which are almost the same by statement and have very little difference.
Complete step by step solution:
Remainder theorem is the theorem in which when we have any polynomial f(x) and we divide it by the binomial (x-a) then the remainder that we get after dividing by the binomial is equal to f(a).
Now, in the factor theorem if a is the root of the polynomial f(x) then we can say that (x-a) is a factor of f(x) and conversely, if (x-a) is a factor of f(x) then we can say that a is the zero of the polynomial f(x).
Let us see both the theorems using an example:
$f\left( x \right)={{x}^{2}}-2x+1$
Firstly, using the remainder theorem let us put 3 in the place of x in f(x),
Then,
$\begin{align}
& f\left( 3 \right)={{3}^{2}}-2\left( 3 \right)+1 \\
& \Rightarrow f\left( 3 \right)=9-6+1 \\
& \Rightarrow f\left( 3 \right)=4 \\
\end{align}$
Therefore, we can say that using remainder theorem if we divide the given polynomial $f\left( x \right)={{x}^{2}}-2x+1$by the binomial $\left( x-3 \right)$ we get the remainder as 4.
Similarly, now applying the factor theorem on the same polynomial.
Now, we can see that the polynomial taken is quadratic so it is zero
$\begin{align}
& f\left( 1 \right)={{1}^{2}}-2\times 1+1 \\
& \Rightarrow 2-2 \\
& \Rightarrow 0 \\
\end{align}$
when x=1 and this implies that (x-1) is the factor of the taken quadratic polynomial.
And we see conversely, of the factor theorem then we can see that if we factor the given quadratic polynomial, we get ${{x}^{2}}-2x+1={{\left( x-1 \right)}^{2}}$which clearly states that 1 is the repeated root of the taken quadratic polynomial.
Note: Be careful while applying the remainder theorem and the factor theorem as the statement of both these theorems sometimes seems to be same and hence lead to wring results. Apart from this try to make the factors wisely and do the converse correctly.
Complete step by step solution:
Remainder theorem is the theorem in which when we have any polynomial f(x) and we divide it by the binomial (x-a) then the remainder that we get after dividing by the binomial is equal to f(a).
Now, in the factor theorem if a is the root of the polynomial f(x) then we can say that (x-a) is a factor of f(x) and conversely, if (x-a) is a factor of f(x) then we can say that a is the zero of the polynomial f(x).
Let us see both the theorems using an example:
$f\left( x \right)={{x}^{2}}-2x+1$
Firstly, using the remainder theorem let us put 3 in the place of x in f(x),
Then,
$\begin{align}
& f\left( 3 \right)={{3}^{2}}-2\left( 3 \right)+1 \\
& \Rightarrow f\left( 3 \right)=9-6+1 \\
& \Rightarrow f\left( 3 \right)=4 \\
\end{align}$
Therefore, we can say that using remainder theorem if we divide the given polynomial $f\left( x \right)={{x}^{2}}-2x+1$by the binomial $\left( x-3 \right)$ we get the remainder as 4.
Similarly, now applying the factor theorem on the same polynomial.
Now, we can see that the polynomial taken is quadratic so it is zero
$\begin{align}
& f\left( 1 \right)={{1}^{2}}-2\times 1+1 \\
& \Rightarrow 2-2 \\
& \Rightarrow 0 \\
\end{align}$
when x=1 and this implies that (x-1) is the factor of the taken quadratic polynomial.
And we see conversely, of the factor theorem then we can see that if we factor the given quadratic polynomial, we get ${{x}^{2}}-2x+1={{\left( x-1 \right)}^{2}}$which clearly states that 1 is the repeated root of the taken quadratic polynomial.
Note: Be careful while applying the remainder theorem and the factor theorem as the statement of both these theorems sometimes seems to be same and hence lead to wring results. Apart from this try to make the factors wisely and do the converse correctly.
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