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# If (x3+ax2−bx+10) is divisible by x2−3x+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 (x−a) is the factor of the polynomial p(x). Let p(x)=x3+ax2−bx+10 and g(x)=x2−3x+2 Factorise g(x)=x2−3x+2:x2−3x+2=x2−2x−x+2=x(x−2)−1(x−2)=(x−2)(x−1)Therefore, g(x)=(x−2)(x−1)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=a−b+11p(2)=23+(a×22)−(b×2)+10=8+(a×4)−2b+10=4a−2b+18Now equate p(2)=0 and p(1)=0 as shown below:a−b+11=0⇒a−b=−11.......(1)4a−2b+18=0⇒2(2a−b+9)=0⇒2a−b+9=0⇒2a−b=−9.......(2)Now subtract equation 1 from equation 2: (2a−a)+(−b+b)=(−9+11)⇒a=2Substitute a=2 in equation 1:2−b=−11⇒−b=−11−2⇒−b=−13⇒b=13Hence, a=2 and b=13.

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