Question

When kx^3 + px^2 - x + 3 is divided by x – 1, the remainder is 4. When kx^3 + px^2 - x + 3 is divided by x - 2, the remainder is 21. How can I find the values of k and p?

Answer 1

if f(x) = kx^3 – px^2 - x + 3
then f(1) will be the remainder when divided by (x-1)

Answer 2

I'm afraid I don't understand your answer ranga. F(1)=k(1)^3-p(1)^2-(1)+3 still leaves me with an undiscovered value for both k and p, unless I am misunderstanding your response.

Answer 3

f(x) = kx^3 – px^2 - x + 3
f(1) = k - p - 1 + 3 = k - p + 2
f(1) = remainder when f(x) is divided by (x-1)
Therefore, k - p + 2 = 4
k - p = 2 ---- (equation 1)

g(x) = kx^3 + px^2 - x + 3
do the same.
g(2) will be the remainder when divide by (x-2)
You will get the second equation.
Two equations and two unknowns k and p.
solve for k and p.

Answer 4

g(2)=k(2)^3-p(2^2)-(2)+3
g(2)=k8-p4+1
21-1=k8-p4+1-1
20=k8-p4

So, now I've got k-p=2 and k8-p4=20

I think I'm beginning to understand but I'm still stuck at the part where simply having two different equations is going to help solve this. In both equations (k-p=2 and k8-p4=20) there are the unsolved values of k and p and no way that I can find to solve either so I can finish either equation.

I just don't understand how I can get k-p=2 and k8-p4=20 to give me solid answers for either of the variables that don't involve some permutation of k or p equaling some for of each other (i.e. k=2+p, etc.)

Answer 5

I should also add that I made a mistake when I wrote the question. The final equations I'm working with are k+p=2 and k8+p4=20, rather than the written minus signs in previous responses.

Answer 6

k + p = 2 --- (1)
8k + 4p = 20 ---- (2)
multiply (1) by 4:
4k + 4p = 8. Subtract this from (2)
4k + 0 = 12
divide by 4
k = 3
put k = 3 in (1):
3 + p = 2
p = 2 -3 = -1

k = 3; p = -1

Answer 7

Oh! I see now! I didn't even think of multiplying the equation to be able to subtract one from the other and isolate one of the variables. I feel so stupid now T.T

Thank you so much for your help ranga! I really do appreciate it.

Answer 8

You are very welcome.

Answer 9

You could also solve the system by substitution:

\[k+p=2\]\[k=2-p\]\[8k+4p=20\]\[8(2-p)+4p=20\]\[16-8p+4p=20\]\[-4p=4\]\[p=-1\]\[k=2-(-1) = 3\]

Answer 10

and testing the original problem:

\[kx^3 + px^2 - x + 3 = 3x^3-x^2-x+3\]\[3(1)^3-(1)^2-(1)+3 = 3-1-1+3=4\checkmark\]
\[3(2)^3-(2)^2-(2)+3=24-4-2+3=21\checkmark\]

(We're making use of the remainder theorem here)

Answer 11

Oh god I feel even more stupid now! Thank you both so much, honestly. These answers seem so obvious now that I seem them written out (especially yours whpalmer) but it's hard to get around those mental blocks some times. I'll make sure I remember your method palmer, it should help a lot.

My only regret is that I can't give you both medals. Cheers!