Can anyone help to solve the problem circled?
- Thread starter yovaraj
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D
Deleted member 4993
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You have been asked to show your work in your previous post - we are still waiting! For this problem, I would start with remainder theorem.
Please show us what you have tried and exactly where you are stuck.
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Please share your work/thoughts about this problem.
HallsofIvy
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Saying that \(\displaystyle 2x^3+ ax^2- 10x+ b\) has x- 1 as a factor means we can write \(\displaystyle 2x^3+ ax^2- 10x+ b= (x-1)Q(x)\) where Q(x) is some quadratic polynomial. And that, in turn, means that if x= 1, \(\displaystyle 2(1)^3+ a(1)^2- 10(1)+ b= a+ b- 8= 0\) or a+ b= 8.
Saying that \(\displaystyle 2x^3+ ax^2- 10x+b\) "remainder -8 when divided by x- 2" meas that we can write \(\displaystyle 2x^3+ ax^2- 10x+ b= (x-2)R(x)- 8\) where R(x) is some quadratic polynomial. And that, in turn, means that when x= 2, \(\displaystyle 2(2)^3+ a(2)^2- 10(2)+ b= 16+ 4a- 20+ b= -8\) so 4a+ b- 4= -8 or 4a+ b= -4.
Solve the equations a+ b= 8 and 4a+ b= -4.
Saying that \(\displaystyle 2x^3+ ax^2- 10x+b\) "remainder -8 when divided by x- 2" meas that we can write \(\displaystyle 2x^3+ ax^2- 10x+ b= (x-2)R(x)- 8\) where R(x) is some quadratic polynomial. And that, in turn, means that when x= 2, \(\displaystyle 2(2)^3+ a(2)^2- 10(2)+ b= 16+ 4a- 20+ b= -8\) so 4a+ b- 4= -8 or 4a+ b= -4.
Solve the equations a+ b= 8 and 4a+ b= -4.
D
Deleted member 4993
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Prof. Ivey,
I suggest we wait till the OP shows a modicum of effort - prior to us providing "solution".