If (x^2-x-2) is HCF of the polynomials (x-2)(2x^2+ax+1) and (x+1)(3x^2+bx+2), find the value of a.

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25 Jul 2013 HCF

If (x^2-x-2) is HCF of the polynomials (x-2)(2x^2+ax+1) and (x+1)(3x^2+bx+2), find the value of a.

OPtion

1) 1
2) 2
3) 3
4) 4
5) 5
6) 6
7) 7
8) 8
9) 9
10)None of these

Solution

Let P(x)=(x-2)(2x^2+ax+1) and Q(x)=(x+1)(3x^2+bx+2)
Factorize (x^2-x-2)=(x-2)(x+1)
As (x-2)(x+1) is the HCF of P(x) and Q(x), both (x-2) and (x+1) must be factor of P(x) and Q(x)
Solve
a=3
option

Correct Option: 3 Best Solution (16)

chandrashekhar kumar   10 years AGO

Factor of (x^2-x-2) is (x+1)(x-2)
SO from 1st equation (x-2)(2x^2+ax+1) we need to get (x+1). SO if a=3 then its esily find.

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mohan lal garg   10 years AGO

(x^2-x-2) =(x-2)(x+1)
is hcf so it term is necessary in both polynomial
so
(x-2)(2x^2+ax+1)=(x-2)(x+1)(2x+c)
at x=-1
2x^2+ax+1=0
2-a+1=0
a=3
ans is 3

Like? Yes (3) | No   

Jigyasa Kumar   10 years AGO

a = 3 satisfies the given condition

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Aarush Singh   10 years AGO

a= 3 satisfies the condition

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anand s idagal   10 years AGO

x^2-x-2..we will get factors as x=-1,+2
nw,substitute x=-1 in (x-2)(2x^2+ax+1) we get a=3

Like? Yes (2) | No   

sajjan kumar   10 years AGO

x^2-x-2=(x-2)(x+1)
2x^2+ax+1 should have a factor (x+1)
so a=3
it will be (x+2)(x+1)

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Subesh Kumar Yadav   10 years AGO

put the value of the a and check by option

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raghu   10 years AGO

(x^2-x-2)=(x+1)(x-2)
divide with this
(x-2)(2x^2+ax+1)/(x+1)(x-2)=(2x^2+ax+1)/(x+1)
substitute each and check 3 is valid

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Shekhar Suman   10 years AGO

ans. 3)3
(x+1)(x-2) divide ((x-2)(2x^2+ax+1))then result is 0

so
x-1 divide (2x^2+ax+1) is 0
now put x=1
(2.1^2+a.1+1) = 0
a=3

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ravi singh   10 years AGO

(x+1)*(x-2) is a factor then they satisfy it by putting x=-1 we get a=3

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shekhar pankaj   10 years AGO

Given x^2 -x-2=(x-2)(x+1) is an hcf of given two polynomials
so (x-2)(2x^2+ax+1)is divisible by(x-2)(x+1)
or (2x^2+ax+1)is divisible by (x+1)

for q=2x+1
which give remainder as a-2-1 =0
so a=3

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PREM KUMAR   10 years AGO

x^2-x-2 as two factor x-2 and x+1 and so from HCF property both polynomial should have only these two factor has common.
In first polynomial x-2 is already there so getting x+1 as a factor a will have value 3.

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Gaurav Sharma   10 years AGO

since x^2-x-2 can be factored to (x-2)(x+1)
since (x-2)(2x^2+ax+1) already contains(x-2) and to be divisible by x^2-x-2 it must also be divisible by x+1 so we chose a as it gives a (x+1)as a factor so we take 3 hence answer

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Ann Mary Jolly   10 years AGO

roots of HCF = -1,2
They will also be the roots of the equations.
Substituting x = -1 in first equation,
we get a = 3

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Vinay Sipani    10 years AGO

(x^2 - x - 2) = (x - 2)(x + 1)

=> One factor of (2x^2 + ax + 1) is (x + 1)
=> -a/2 = -1 - 1/2 => a = 3

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aniruddh shrivastava   10 years AGO

see, instead of solving equations we can find out the answer by just putting the values....
suppose the value of x is 4,
now HCF= x^2-x-2 = 16-4-2=10, HCF = 10...
since we have to determine the value of a then we can ignore second polynomial....
putting the value of x=4 in 1st polynomial we get
(2)(32+4a+1)= 66+8a
now we can predict that 66+8a should be a multiple of 10(HCF)....then we can put the value of a from the given options..... if we put 3 => 66+8a=>66+24=>multiple of 10.....so 3 is the right option...

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