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Question

If x3+ax2bx+10 is exactly divisible by x2+3x+2. Find the values of a and b

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Solution

x2+3x+2=(x+2)(x+1))
Since f(x) is divisible by (x+2)(x+1)
f(2)&(f1) must be zero \rightarrow Remainder theorem
f(2)=(2)3+a(2)2b(2)+10=08+4a+2b+10=0
or 4a+2b+2=0 -----(1)
f(1)=(1)3+a(1)2b(1)+10=0=1+a+b+10=0a+b+9=0 -----(2)
Solving (1)&(2)
put a=(b+a)is(1)4(b+9)+2b+2=04b36+2b+2=02b34=0orb=17a=(17+9)=26

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