We need divide f(x)=2x4−4x3−3x−1 by (x−1).
Procedure:
Step 1:
2x4x=2x3
Now multiply (x−1)(2x3)=2x4−2x3 and then subtract.
Step 2:
−2x3x=−2x2
Now multiply (x−1)(−2x2)=−2x3+2x2 and then subtract.
Step 3:
−2x2x=−2x
Now multiply (x−1)(−2x)=−2x2+2x and then subtract.
Step 4:
−5xx=−5
Now multiply (x−1)(−5)=−5x+5 and then subtract.
Here the quotient is 2x3−2x2−2x−5 and the remainder is −6.Now, the zero of the polynomial (x−1) is 1.Put x=1 in f(x),f(x)=2x4−4x3−3x−1f(1)=2(1)4−4(1)3−3(1)−1=2(1)−4(1)−3(1)−1=2−4−3−1=−6Is the remainder same as the value of the polynomial f(x) at zero of (x−1)?From the above examples we shall now state the fact in the form of the following theorem.It gives a remainder without actual division of a polynomial by a linear polynomial in one variable.
Given polynomial is f(x)=2x4−4x3−3x−1 and divided by (x−1)
Put x=1 in the given polynomial, we get
f(x)=2x4−4x3−3x−1
⇒f(x)=2(1)4−4(1)3−3(1)−1
⇒f(x)=2−4−3−1
⇒f(x)=−6
Then 2x4−4x3−3x−1 divided by (x−2) the reminder is −6.
Then polynomial 2x4−4x3−3x−1 divided by (x−2) the reminder is not zero.