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if 2y³+ay²-11y+b leaves remained 0 and 42 when divided by (y-2) and (y-3) respectively, find the valu of a and b, with the value of aand b, factorise the given expression. |
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Answer» Answer: Then, x=2 Given, f(x)=2x 3 +ax 2 –11x+b Now, substitute the value of x in f(x), f(2)=2(2) 3 +a(2) 2 –11(2)+b =16+4a–22+b =–6+4a+b Given, remainder is 0. So,−6+4a+b=0 4a+b=6 … [equation (i)] Now, CONSIDER (x – 3) Assume x–3=0 Then, x=3 Given,f(x)=2x 3 +ax 2 –11x+b Now, substitute the value of x in f(x), f(2)=2(3) 3 +a(3) 2 –11(3)+b =54+9a–33+b =21+9a+b Given, remainder is 42. So, 21+9a+b=42 9a+b=42–21 9a+b=21 … [equation (II)] Now, subtracting equation (i) from equation (ii) we get, (9a+b)–(4a+b)=21–6 9a+b–4a–b=15 5a=15 a=15/5 a=3 Consider the equation (i) to find out ‘b’. 4a+b=6 4(3)+b=6 12+b=6 b=6–12 b=−6 Then, by substituting the value of a and bf(x)=2x 3 +3x 2 –11x–6 Given that remainder is 0 for, (x–2) is a factor of f(x). So, DIVIDING f(x) by (x–2) Therefore, 2x 3 +3x 2 –11x–6=(x–2)(2x 2 +7x+3) =(x–2)(2x 2 +6x+x+3) =(x–2)(2x 2 +6x+x+3) =(x–2)(2x(x+3)+1(x+3)) =(x–2)(x+3)(2x+1) |
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