03Dremainder_factor

The Remainder Theorem
For any polynomial P(x)

If we divide P(x) by (x – a) then

The ** __remainder__ ** will be the same as P(a)


 * Example 1 **

Given the polynomial ... ... P(x) = x 2 + 5x + 2

Is divided by (x + 1).

Find the remainder.


 * Solution:**

... ... The result of performing the long division is shown on the right.

... ... It can be seen that the remainder is –2.

OR

... ... P(–1) = (–1) 2 + 5(–1) + 2

... ... P(–1) = 1 – 5 + 2

... ... P(–1) = –2

... ... According to the Remainder Theorem

... ... The remainder will be –2

The Factor Theorem

 * If (x – a) is a factor of P(x)
 * A factor is a term that divides evenly into the original expression.
 * P(x) ÷ (x – a) will have a remainder of 0.
 * Putting this together with the Remainder Theorem, we have

For any polynomial, P(x)

If P(a) = 0 then

(x – a) is a factor of P(x)


 * Example 2 **

Confirm whether or not (x + 2) is a factor of the polynomial P(x) = x 3 + 5x 2 – 2x + 3


 * Solution:**

... ... P(x) = x 3 + 5x 2 – 2x + 3

... ... P(–2) = (–2) 3 + 5(–2) 2 – 2(–2) + 3

... ... P(–2) = –8 + 20 + 4 + 3

... ... P(–2) = 19

... ... P(–2) is __not__ equal to 0

... ... so (x + 2) is __not__ a factor of P(x)


 * Finding Factors of a Polynomial **


 * The factor theorem is useful in finding factors of a polynomial
 * Try different values of a until P(a) = 0
 * (x – a) is then a factor.


 * Example 3 **

Find a factor of P(x) = x 3 + 3x 2 + x – 2

Solution:

... ... P(x) = x 3 + 3x 2 + x – 2

... ... P(1) = (1) 3 + 3(1) 2 + (1) – 2 = 3

... ... P(–1) = (–1) 3 + 3(–1) 2 + (–1) – 2 = –1

... ... P(2) = (2) 3 + 3(2) 2 + (2) – 2 = 20

... ... P(–2) = (–2) 3 + 3(–2) 2 + (–2) – 2 = 0

... ... P(–2) = 0 so (x + 2) is a factor.




 * The Factor Theorem on the Classpad Calculator **


 * This repeated use of the same polynomial is an ideal use of the define command
 * define is in the ACTION menu, COMMAND submenu
 * use p from the ABC tab of the virtual keyboard
 * use //**x **// from the keypad

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