Use Remainder Theorem to solve polynomial equations

In division numbers, we often get things like the following example. At division 17: 5 is 3 with the remaining 2.
17 = (5 x 3) + 2

The numbers are 17 as divided numbers, 5 as dividers, 3 as divides, and 2 as the remainder.
By looking at the similarities above, we can also determine the basic equation that connects the polynomial f(x) which in this case is a divided element. P(x) as a divider, H(x) as a result of divide and S is the remainder of the division, i.e.

f(x) = P(x) . H(x) + S

for P(x) = x – k, then
f(x) = (x – k) . H(x) + S

Based on the above equation, we can mention a theorem for the remainder of the polynomial division.

Polynomial Remainder Theorem

If the Polynomial f(x) is divided by x - k, the remainder is f (k).

Proof:
Use the equation f(x) = (x – k) . H(x) + S.
The degree S is lower than x - k, therefore S is a constant.
For x = k, it will be obtained
f(k) = (k – k) . H(k) + S
f(k) = 0. H(k) + S

so, f(k) = S (proven).

Example 1.
Determine the quotient and remainder of the division of the polynomial equation 3x4 – 2x3 + x – 7 divided by x - 2.


Then the results obtained are 3x3 + 4x2 + 8x + 17 with the remaining 27.

We try to multiply between (x - 2) with the results obtained previously, the result is the initial equation before divided (x - 2).

(x – 2)( 3x3 + 4x2 + 8x + 17) + 27
3x4 + 4x3 + 8x2 + 17x – 6x3 – 8x2 – 16x - 34 + 27
3x4 + 4x3– 6x3 + 8x2– 8x2+ 17x – 16x - 34 + 27
3x4 – 2x3 + x + 61.

The Polynomial Equation Is Divided By ax - b

Using the remainder theorem above can determine a new equation for the polynomial f (x) divided by ax - b, as follows.

f(x) = (ax – b) H(x) + S
f(x) = a(x – b/a) H(x) + S
f(x) = (x – b/a) . a H(x) + S

In the above equation, it can be shown that the remainder of the polynomial division f(x) by ax - b is f(b/a).

Example 2.
Determine the quotient and remainder of the division of the polynomial equation 
3x3 + 5x2 – 11x + 8 divided by 3x – 1.


f(x) = (x – 1/3) (3x2 + 6x – 9) + 5
f(x) = (x – 1/3) . 3 . (x2 + 2x – 3) + 5
f(x) = (3x – 1)( x2 + 2x – 3) + 5

Then the results obtained are x2 + 2x – 3 with the remaining 5.

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