When x^4 + ax^3 + bx +c is divided by (x–1), (x+1), and (x+2), the remainders are 13, 3, and –12 respectively. Find the values of a, b, and c
+154 The remainder theorem tells us that:
If a polynomial f(x) is divided by (x-a), then the remainder is f(a).
So for this problem, the polynomial is divided by (x-1), (x+1), and (x+2) and the remainders are given.
Therefore, we can conclude that
f(1) = 13
f(-1) = 3
f(-2) = -12
You can now plug in those x values and y values to solve for a, b, and c.
\(x^4+ax^3+bx+c\\ 1^4+a(1)^3+b(1)+c=13\implies a+b+c=12\\ (-1)^4+a(-1)^3+b(-1)+c=3\implies -a-b+c=2\\ (-2)^4+a(-2)^3+b(-2)+c=-12\implies -8a-2b+c=-28\)
We now have a system of three equations.
Now to find a, b, and c.
We can first add the first two equations together
\(a+b+c=12\\ -a-b+c=2\\ ==========\\ 2c=14\\ c=7 \)
We can then multiply the 2nd equation by -2 and add it with the 3rd equation
\(\underset{\downarrow}{-2(-a-b+c)=-2(2)}\\ 2a+2b-2c=-4\\ -8a-2b+c=-28\\ ============\\ -6a-c=-32\\ -6a-7=-32\\ -6a=-25\\ a=\frac{25}{6} \)
Now, we can plug in a and c into any other equation to find b.
\(a+b+c=12\\ \frac{25}{6}+b+7=12\\ \frac{25}{6}+b=5\\ b=\frac{5}{6}\)
TL;DR
a = 25/6, b = 5/6, c = 7
