The polynomials ax + 3x - 3 and 2x - 5x + a, when divided by (x - 4) leaves remainders
R, & R, respectively then value of a if 2R, -R, 0:
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Answer:
Let f(x) = x4 – 2x3 + 3x2 – ax + b
Now,
f(1) = 14 – 2(1)3 + 3(1)2 – a(1) + b
5 = 1 – 2 + 3 – a + b
3 = - a + b (i)
And,
f(-1) = (-1)4 – 2(-1)3 + 3(-1)2 – a(-1) + b
19 = 1 + 2 + 3 + a + b
13 = a + b (ii)
Now,
Adding (i) and (ii),
8 + 2b = 24
2b = 16
b = 8
Now,
Using the value of b in (i)
3 = - a + 8
a = 5
Hence,
a = 5 and b = 8
Hence,
f(x) = x4 – 2(x)3 + 3(x)2 – a(x) + b
= x4 – 2x3 + 3x2 – 5x + 8
f(2) = (2)4 – 2(2)3 + 3(2)2 – 5(2) + 8
= 16 – 16 + 12 – 10 + 8
= 20 – 10
= 10
Therefore, remainder is 10..
Step-by-step explanation:
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