Pls solve this 5th question. ..I really need it's solution
.
.....do favour me
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Answered by
1
hope it helps you......................................................................................................................
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Akanksha1710:
thanks for solving my question and giving your auspicious time to it but ......I have to say dear that your answers are wrong maybe there is some mistakes in your calculations....do check them and again thank u very much ..
yes!!i forget X^3
ed*
nothing to worry ....every person makes mistake. .it's common dear
ok!!
thanks.!!
next time i will take care of it...
Answered by
3
Hiii friend,
(X+1) and (X-2) are the two factor of the given polynomial.
(X+1) = 0 and X -2 = 0
X = -1 and X = 2
P(X) = X³+(A+1)X² -(B-2)X-6
P(-1) = (-1)³ + (A+1) × (-1)² - (B-2) × - 1 -6
=> -1 + A +1 - (-B+2) -6
=> -1 + A + 1 + B -2 -6
=> A+B -8 = 0
=> A + B = 8.......(1)
Also X = 2
P(2) = (2)³ + (A+1) × (2)² - (B-2) × 2 -6
=> 8 + (A+1) × 4 -(2B -4) -6
=> 8 + 4A + 4 -2B +4 -6
=> 4A - 2B +10 = 0
=> 4A - 2B = -10.......(2)
From equation 1 we get,
A + B = 8
A = 8-B......(3)
Putting the value of A in equation (2)
4A - 2B = -10
4 × (8-B) -2B = -10
32 - 4B -2B = -10
32 - 6B = -10
-6B = -10-32
-6B = -42
B = -42/-6 = 7
Putting the value of B in equation (3)
A = 8-B
A = 8 -7 = 1.
Hence,
A = 1 and B = 7
HOPE IT WILL HELP YOU..... :-)
(X+1) and (X-2) are the two factor of the given polynomial.
(X+1) = 0 and X -2 = 0
X = -1 and X = 2
P(X) = X³+(A+1)X² -(B-2)X-6
P(-1) = (-1)³ + (A+1) × (-1)² - (B-2) × - 1 -6
=> -1 + A +1 - (-B+2) -6
=> -1 + A + 1 + B -2 -6
=> A+B -8 = 0
=> A + B = 8.......(1)
Also X = 2
P(2) = (2)³ + (A+1) × (2)² - (B-2) × 2 -6
=> 8 + (A+1) × 4 -(2B -4) -6
=> 8 + 4A + 4 -2B +4 -6
=> 4A - 2B +10 = 0
=> 4A - 2B = -10.......(2)
From equation 1 we get,
A + B = 8
A = 8-B......(3)
Putting the value of A in equation (2)
4A - 2B = -10
4 × (8-B) -2B = -10
32 - 4B -2B = -10
32 - 6B = -10
-6B = -10-32
-6B = -42
B = -42/-6 = 7
Putting the value of B in equation (3)
A = 8-B
A = 8 -7 = 1.
Hence,
A = 1 and B = 7
HOPE IT WILL HELP YOU..... :-)
thank u very much dear ...I must say you did a great job by writing every step clearly and your answers are right too thank you very much. ..I will consult in future if I have some problems regarding the subject again thank you very much
I am having one more question ..
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