Find the value 'a' for which x^3-7x+5 is a factor of x^5-2x^4-4x^3+19x^2-31x+21+a.
Answers
Answer:
Answer
Given that x
3
−7x+5 is a factor of x
5
−2x
4
−4x
3
+19x
2
−31x+12+a
That is the remainder after division should be 0
Let us first divide x
5
−2x
4
−4x
3
+19x
2
−31x+12+a by x
3
−7x+5 using long division method
x
2
−2x+3
x
3
−7x+5)
x
5
−2x
4
−4x
3
+19x
2
−31x+12+a
x
5
+0x
4
−7x
3
+5x
2
(−)(−)(+)(−)
−2x
4
+3x
3
+14x
2
−31x
−2x
4
+0x
3
+14x
2
−10x
(+)(−)(−)(+)
3x
3
−21x+12+a
3x
3
−21x+15
(−)(+)(−)
−3+a
Here the remainder is −3+a
But as x
3
−7x+5 is a factor of x
5
−2x
4
−4x
3
+19x
2
−31x+12+a
∴−3+a=0
∴a=3.