Using the Remainder Theorem, we know that the remainder when k(x) is divided by x-1 is k(1)=14 and the remainder when k(x) is divided by x-2 is k(2)=55.
Let's use synthetic division to divide k(x) by x-1:
1 | 2 0 a b 3
--|-------
| 2 2a+2b+a+2 3+a+b
--------------------
2a+2b+a-1 14-a-b
Since the remainder is 14, we know that 2a+2b+a-1 = 14-a-b. Simplifying, we get:
3a+3b=15
a+b=15/3
a+b=5
Now let's use synthetic division to divide k(x) by x-2:
2 | 2 0 a b 3
--|--------
| 4 8a+4b+4 16a+8b+12
--------------------
2a+4b-16 55+16a+8b
Since the remainder is 55, we know that 2a+4b-16 = 55+16a+8b. Simplifying, we get:
14a+4b=-39
7a+2b=-39/2
We can solve these two equations simultaneously by multiplying the second equation by 2 and subtracting it from the first equation:
14a+4b-7a-2b=-39-(-39)
7a+2b=0
Solving for b in terms of a, we get b=-7a/2. Substituting into the equation a+b=5, we get:
a-7a/2=5
a=10/3
Substituting a=10/3 into b=-7a/2, we get:
b=-35/3
Therefore, the values of a and b are a=10/3 and b=-35/3.
Find the value of a and b if k(x)=2x^4+ax^2+bx+3 is divided by x-1the remainder is 14 and When k(x) is divided by x-2, the reminder is 55.
3 answers
AAAaannndd the bot gets it wrong yet again!
k(x)=2x^4+ax^2+bx+3
k(1)=14, k(2) = 55, so
2+a+b+3 = 14
32+4a+2b+3 = 55
That gives a=1, b=8.
Hence,
k(x)=2x^4+x^2+8x+3
k(x)=2x^4+ax^2+bx+3
k(1)=14, k(2) = 55, so
2+a+b+3 = 14
32+4a+2b+3 = 55
That gives a=1, b=8.
Hence,
k(x)=2x^4+x^2+8x+3
Apologies for the error in my previous response. You are correct that the correct values of a and b are a=1 and b=8, and the polynomial k(x) is:
k(x) = 2x^4 + x^2 + 8x + 3.
Thank you for pointing out the mistake.
k(x) = 2x^4 + x^2 + 8x + 3.
Thank you for pointing out the mistake.