38 MATHEMATICS
Let us consider some more examples.
Example 7 : Divide the polynomial 3x^4 – 4x^3 – 3x –1 by x – 1.
Solution : By long division, we have:
3 x^3 – x^2 – x – 4
x – 1 3 x (^4) – 4x (^3) – 3x – 1
(^) – 3 x^4 –+ 3 x^3
- x^3 – 3 x – 1
x^3 +
x^2
- x^2 – 3x – 1
- x^2 +– x
- 4x – 1
4 x +
4
– 5
Here, the remainder is – 5. Now, the zero of x – 1 is 1. So, putting x = 1 in p(x), we see
that
p(1) = 3(1)^4 – 4(1)^3 – 3(1) – 1
= 3 – 4 – 3 – 1
= – 5, which is the remainder.
Example 8 : Find the remainder obtained on dividing p(x) = x^3 + 1 by x + 1.
Solution : By long division,
x^2 – x + 1
x + 1 x^3 + 1
(^) – x^3 +– x^2
- x^2 + 1
x^2 –
+
x
x + 1
x +