If r and s be the remainder when the polynomials
(x³+2x²-5ax-7) and (x³+ax²-12ax+16 are divy by (x+1) and (x+2) respectively and 2r +s=6
Then find the value of a =?
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Hi there!
P(x) = x³+2x²-5ax-7
Putting the value of x =-1 from (x+1)
P (-1) = (-1)³+2(-1)²-5a(-1) -7
= -1+ 2+ 5a- 7
= -6+5a
Q(x) = x³+ax²-12a+6
Putting the value of x = 2 from (x-2)
Q(2) = (2)³+ a(2)²- 12(2) +6
= 8+ 4a- 24+ 6
= -10+ 4a
Given that,
2r + s= 6
2(-6+5a) + (-10+4a) = 6
-12+10a-10+4a = 6
-22+ 14a = 6
14a = 6+ 22
a = 28/14
a = 2
Therefore,
Value of 'a' is 2.
[ Thank you! for asking the question. ]
Hope it helps!
P(x) = x³+2x²-5ax-7
Putting the value of x =-1 from (x+1)
P (-1) = (-1)³+2(-1)²-5a(-1) -7
= -1+ 2+ 5a- 7
= -6+5a
Q(x) = x³+ax²-12a+6
Putting the value of x = 2 from (x-2)
Q(2) = (2)³+ a(2)²- 12(2) +6
= 8+ 4a- 24+ 6
= -10+ 4a
Given that,
2r + s= 6
2(-6+5a) + (-10+4a) = 6
-12+10a-10+4a = 6
-22+ 14a = 6
14a = 6+ 22
a = 28/14
a = 2
Therefore,
Value of 'a' is 2.
[ Thank you! for asking the question. ]
Hope it helps!
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