If p and q are remainders when the polynomials x3 + 2x2 -5ax -7 and x3 + ax2 -12x + 6 are divided by x + 1 and x -2 respectively and if 2p + q = 6 , find a.
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Answers
EXPLANATION.
P are remainders when the polynomials,
x³ + 2x² - 5ax - 7 are divided by (x + 1).
q are the remainders when the polynomials,
x³ + ax² - 12x + 6 are divided by (x - 2).
As we know that,
⇒ (x + 1) = 0.
⇒ x = - 1.
Put the value of x = - 1 in the equation, we get.
⇒ x³ + 2x² - 5ax - 7 = p.
⇒ (-1)³ + 2(-1)² - 5a(-1) - 7 = p.
⇒ - 1 + 2 + 5a - 7 = p.
⇒ 2 + 5a - 8 = p.
⇒ 5a - 6 = p. - - - - - (1).
⇒ (x - 2) = 0.
⇒ x = 2. - - - - - (2).
Put the value of x = 2 in the equation, we get.
⇒ x³ + ax² - 12x + 6 = q.
⇒ (2)³ + a(2)² - 12(2) + 6 = q.
⇒ 8 + 4a - 24 + 6 = q.
⇒ 4a - 24 + 6 + 8 = q.
⇒ 4a - 24 + 14 = q.
⇒ 4a - 10 = q. - - - - - (2).
It is given that,
⇒ 2p + q = 6.
Put the values of p and q in the equation, we get.
⇒ 2(5a - 6) + (4a - 10) = 6.
⇒ 10a - 12 + 4a - 10 = 6.
⇒ 10a + 4a - 12 - 10 = 6.
⇒ 14a - 22 = 6.
⇒ 14a = 6 + 22.
⇒ 14a = 28.
⇒ a = 2.
Values of a = 2.
Given :-
If p and q are remainders when the polynomials x³ + 2x² - 5ax - 7 and x³ + ax² -12x + 6 are divided by x + 1 and x -2 respectively and if 2p + q = 6 ,
To Find :-
Value of a
Solution :-
x + 1 = 0
x = 0 - 1
x = -1
Putting x = -1
p = (-1)³ + 2(-1)² - 5a(-1) - 7
p = -1 + 2(1) - (-5a) - 7
p = -1 + 2 + 5a - 7
p = 5a + 2 - 8
p = 5a - 6
Now
x - 2 = 0
x = 0 + 2
x = 2
Putting x = 2
q = (2)³ + a(2)² - 12(2) + 6
q = 8 + a(4) - 24 + 6
q = 8 + 4a - 24 + 6
q = 4a - 24 + 14
q = 4a - 10
Now
2p + q = 6
2(5a - 6) + 4a - 10 = 6
10a - 12 + 4a - 10 = 6
14a - 22 = 6
14a = 22 + 6
14a = 28
a = 28/14
a = 2
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