Online Remainder Theorem, Long Division Polynomials Homework Help
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REMAINDER THEOREM:
Dividend = (Divisor × Quotient) + Remainder
If p(x) and t(x) are two polynomials such that degree of p(x) ≥ degree of t(x) and t(x) ≠ 0, then polynomials q(x) and r(x) can be found such that:
p(x) = t(x) q(x) + r(x),
Where :
r(x) = 0 or degree of r(x) < degree of t(x)
Or
It can be said that p(x) divided by t(x) gives q(x) as quotient and r(x) as remainder.
If p(x) and t(x) are two polynomials such that degree of p(x) ≥ degree of t(x) and t(x) ≠ 0, then polynomials q(x) and r(x) can be found such that:
p(x) = t(x) q(x) + r(x),
Where :
r(x) = 0 or degree of r(x) < degree of t(x)
Or
It can be said that p(x) divided by t(x) gives q(x) as quotient and r(x) as remainder.
Statement of remainder theorem:
"Let p(x) be any polynomial of degree greater than or equal to
one and let a be any real number. If p(x) is divided by the linear
polynomial x – a, then the remainder is p(a)."
Proof:
Let p(x) be any polynomial with degree of p(x) ≥ 1. Assuming that when p(x)
is divided by x- a, the quotient is q(x) and the remainder is r(x), i.e. ,
p(x) = (x – a) q(x) + r(x)
Since the degree of x – a is 1 and the degree of r(x) is less than the degree of q(x), degree of r(x) = 0. This implies that r(a) is a constant, say r.
Hence for every value of x, r(x) = r.
Hence p(x) = (x –a)q(x) + r
If x = a, then the above equation will be:
p(a) = (a – a)q(a) + r
= r (hence proved)
Example:
Find the remainder when x3 + 2x2 + x + 1 is divided by x + 1.
Solution:
Let p(x) = x3 + 2x2 + x + 1
x + 1 = 0 x = -1
Example:
Using long division method, divide the polynomial 3x4 – 4x3 – 3x - 1 by x – 1.
p(x) = (x – a) q(x) + r(x)
Since the degree of x – a is 1 and the degree of r(x) is less than the degree of q(x), degree of r(x) = 0. This implies that r(a) is a constant, say r.
Hence for every value of x, r(x) = r.
Hence p(x) = (x –a)q(x) + r
If x = a, then the above equation will be:
p(a) = (a – a)q(a) + r
= r (hence proved)
Example:
Find the remainder when x3 + 2x2 + x + 1 is divided by x + 1.
Solution:
Let p(x) = x3 + 2x2 + x + 1
x + 1 = 0 x = -1
| Hence remainder | = p(-1) = (-1)3 + 2(-1)2 + (-1) + 1 |
| = -1 + 2 -1 + 1 | |
| = 1 |
Using long division method, divide the polynomial 3x4 – 4x3 – 3x - 1 by x – 1.
Solution:
Hence Quotient = 3x3 - x2 – x – 4 and Remainder = -5.
| 3x3 - x2 – x - 4 | |
| x – 1 | 3x4 – 4x3 – 3x – 1 |
| 3x4 – 3x3 | |
| - + | |
| - x3 – 3x – 1 | |
| - x3 + x2 | |
| + - | |
| -x2 - 3x -1 | |
| -x2 + x | |
| + - | |
| - 4x – 1 | |
| - 4x + 4 | |
| + - | |
| - 5 |
Hence Quotient = 3x3 - x2 – x – 4 and Remainder = -5.
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