Divide this polynomial.

Algebra Level 4

Suppose there exists a polynomial p(x) , which when divided by ( x -1) gives a remainder -1 , when divided by ( x -2) gives a remainder 3 and when divided by ( x +3) gives a remainder 4. Let r(x) be the remainder when that polynomial is divided by ( x -1) ( x -2) ( x +3) . Find the value of r(20) +2.9 .

The answer is 437.

1 solution

Daniel Liu
Apr 8, 2014

From the first few conditions, we know that $p(1)=-1$ , $p(2)=3$ , and (p(-3)=4).

The second part tells us that $p(x)=(x-1)(x-2)(x+3)Q(x)+r(x)$ . We know that $\text{deg}(r(x)) =2$ because $\text{deg}((x-1)(x-2)(x+3))=3$ . Therefore, $r(x)=ax^2+bx+c$ .

Plugging in the values $x=1,2,-3$ in $p(x)=(x-1)(x-2)(x+3)Q(x)+ax^2+bx+c$ gives $p(1)=a+b+c=-1$ $p(2)=4a+2b+c=3$ $p(-3)=9a-3b+c=4$

Solving the system of equations, we arrive at the conclusion that $a=\dfrac{21}{20}$ , $b=\dfrac{17}{20}$ , and $c=-\dfrac{29}{10}$

Therefore, $r(20)+2.9=\dfrac{21}{20}(20)^2+\dfrac{17}{20}(20)-\dfrac{29}{10}+2.9=\boxed{437}$ .

Nice!

Finn Hulse - 7 years, 2 months ago

But why can't r(x) be a linear polynomial or a constabt one?

Adarsh Kumar - 7 years, 1 month ago

If it is linear, solving the system of equation will give $a=0$ .

Shaun Loong - 6 years, 10 months ago

General format for r(x) is ax^2+bx+c. Yes, r(x) can be linear or a constant, if r(x)=0, then a=0 and if r(x)=c (Constant) then a=b=0.

Venkata Karthik Bandaru - 6 years, 3 months ago

Divide this polynomial.

The answer is 437.

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