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Question

When a polynomial $$2x^{3} + 3x^{2} + ax + b$$ is divided by $$(x - 2)$$ leaves remainder $$2$$, and $$(x + 2)$$ leaves remainder $$-2$$. Find $$a$$ and $$b$$.


Solution

The given polynomial is $$P(x)=2x^3+3x^2+ax+b$$ 

It is also given that if $$P(x)$$ is divided by $$(x-2)$$ then it will leave the remainder $$2$$ and if divided by $$(x+2)$$ then the remainder will be $$-2$$ which means that $$P(2)=2$$ and $$P(-2)=-2$$.

Let us first substitute $$P(2)=2$$ in $$P(x)=2x^3+3x^2+ax+b$$ as shown below:

$$P(x)=2x^{ 3 }+3x^{ 2 }+ax+b\\ \Rightarrow P(2)=2(2)^{ 3 }+3(2)^{ 2 }+(a\times 2)+b\\ \Rightarrow 2=(2\times 8)+(3\times 4)+2a+b\\ \Rightarrow 2=16+12+2a+b$$
$$\Rightarrow 2=28+2a+b\\ \Rightarrow 2a+b=2-28\\ \Rightarrow 2a+b=-26\quad .........(1)$$

Now, substitute $$P(-2)=-2$$ in $$P(x)=2x^3+3x^2+ax+b$$ as shown below:

$$P(x)=2x^{ 3 }+3x^{ 2 }+ax+b\\ \Rightarrow P(-2)=2(-2)^{ 3 }+3(-2)^{ 2 }+(a\times -2)+b\\ \Rightarrow -2=(2\times -8)+(3\times 4)-2a+b\\ \Rightarrow -2=-16+12-2a+b$$
$$\Rightarrow -2=-4-2a+b\\ \Rightarrow -2a+b=-2+4\\ \Rightarrow -2a+b=2\quad .........(2)$$

Now adding the equations 1 and 2, we get

$$(2a-2a)+(b+b)=-26+2\\ \Rightarrow 2b=-24\\ \Rightarrow b=-\dfrac { 24 }{ 2 } \\ \Rightarrow b=-12$$

Now substitute the value of $$b$$ in equation 2:

$$-2a+(-12)=2\\ \Rightarrow -2a-12=2\\ \Rightarrow -2a=2+12\\ \Rightarrow -2a=14\\ \Rightarrow a=-\dfrac { 14 }{ 2 } \\ \Rightarrow a=-7$$

Hence $$a=-7$$ and $$b=-12$$

Mathematics

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