Question 1.
Find the remainder when x3 + 3x3 + 3x + 1 is divided by
1. x + 1
2. x – \(\frac {1}{2}\)
3. x
4. x + π
5. 5 + 2x
Solutiobn:
p(x) = x3 + 3x2 + 3x + 1
1. Let
x + 1 = 0
x = – 1
Then p(-l) = (-1)3 + 3(-1)2 + 3(-1) + 1
= -1 + 3 – 3 + 1= 0
∴ Remainder = 0
2. Let x – \(\frac {1}{2}\) = 0
x = \(\frac {1}{2}\)
Then p (\(\frac {1}{2}\)) = (\(\frac {1}{2}\)) + 3(\(\frac {1}{2}\)) + 3\(\frac {1}{2}\) + 1
= \(\frac {1}{8}\) + 3 x \(\frac {1}{4}\) + 3 x \(\frac {1}{2}\) + 1
= \(\frac {1}{8}\) + \(\frac {3}{4}\) + \(\frac {3}{2}\) + 1
= \(\frac{1+6+3 \times 4+1 \times 8}{8}\)
= \(\frac{1+6+12+8}{8}\)
= \(\frac {27}{8}\)
Hence, remainder = \(\frac {27}{8}\)
3. Let x = 0
Then p(0) = o3 + 3(0)2 + 3(0) + 1 = 1
∴ Remainder = 1
4. Let x + π = 0
x = – π
Then p(- π) = (-π)3 + 3(-π)2 + 3π + 1
= -π3 + 3π2 + 3π + 1
∴ Remainder = -π3 + 3π2 + 3π + 1
5. Let 5 + 2x = 0
2x = -5
x = \(\frac {-5}{2}\)
Then p(\(\frac {-5}{2}\)) = (\(\frac {-5}{2}\))3 + 3 (\(\frac {-5}{2}\))2 + 3(\(\frac {-5}{2}\)) + 1
= \(\frac {-125}{8}\) + \(\frac {3 x 25}{4}\) – \(\frac {15}{2}\) + 1
= \(\frac {-125}{8}\) + \(\frac {75}{4}\) – \(\frac {15}{2}\) + 1
= \(\frac{- 125 + 150 – 60 + 8}{8}\) = \(\frac {-27}{8}\)
∴ Remainder = \(\frac {-27}{8}\)
Question 2.
Find the remainder when x3 – ax3 + 6x – a is divided by x – a.
Solution:
Let p(x) = x3 – ax2 + 6x – a
and x – a = 0
= x = a
∴ p(a) = a3 – a x a2 + 6a – a
= a3 – a3 + 6a – a
p(a) = 5a
Hence remainder = 5a
Question 3.
Check whether 7 + 3x is a factor of 3x3 + 7x.
Solution:
7 + 3x will be a factor of polynomial 3x3 + 7x if we divide 3x3 + 7x by 7 + 3x and it leaves no
remainder.
Let p(x) = 3x3 + 7x
and 7 + 3x = 0
3x = – 7
x = \(\frac {-7}{3}\)
Now, p(x) = 3x3 + 7x
So p(\(\frac {-7}{3}\)) = 3(\(\frac {-7}{3}\))3 + 7(\(\frac {-7}{3}\))
= 3 x \(\frac {-343}{9}\) – \(\frac {49}{3}\) = \(\frac {-343 – 147}{9}\) = \(\frac {-490}{9}\)
∴ p(x) = \(\frac {-490}{9}\)
Hence p(x) = 0
∴ 7 + 3x is not a factor of3x3 + 7x.