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Question

If (x3+ax2bx+10) is divisible by x23x+2, find the values of a and b

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Solution

The Factor Theorem states that if a is the root of any polynomial p(x) that is if p(a)=0, then (xa) is the factor of the polynomial p(x).

Let p(x)=x3+ax2bx+10 and g(x)=x23x+2

Factorise g(x)=x23x+2:

x23x+2=x22xx+2=x(x2)1(x2)=(x2)(x1)

Therefore, g(x)=(x2)(x1)

It is given that p(x) is divisible by g(x), therefore, by factor theorem p(2)=0 and p(1)=0. Let us first find p(2) and p(1) as follows:

p(1)=13+(a×12)(b×1)+10=1+(a×1)b+10=ab+11p(2)=23+(a×22)(b×2)+10=8+(a×4)2b+10=4a2b+18

Now equate p(2)=0 and p(1)=0 as shown below:

ab+11=0ab=11.......(1)4a2b+18=02(2ab+9)=02ab+9=02ab=9.......(2)

Now subtract equation 1 from equation 2:

(2aa)+(b+b)=(9+11)
a=2

Substitute a=2 in equation 1:

2b=11
b=112
b=13
b=13

Hence, a=2 and b=13.

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