1. 3.1 The Remainder Theorm
AND
The Factor Theorem

2. Def.: A polynomial function is a function of the form
where n is a nonnegative integer and each ai (i = 0,1,…, n) is a real number.
The polynomial function has a leading
coefficient an and degree n.
Def.:
Polynomial Function

3. Ex1: Divide x2 + 3x – 2 by x – 1 and check the answer.
Dividing Polynomials
Ex1: Divide x2 + 3x – 2 by x – 1 and check the answer. 1. x
+ 2 2.
x x 3. 2x
– 2
2x + 2 4.
– 4 5.
remainder 6.
Answer: x + 2 +
– 4
Check:
(x + 2)
quotient
(x + 1)
divisor
+ (– 4)
remainder
= x2 + 3x – 2
dividend
Dividing Polynomials

4. Notes:
P(x): Dividend
D(x): Divisor
Q(x): Quotient
R(x): Remainder 1)
2) The degree of R(x) <the degree of D(x)

5. Ex2
Ex3(HW)

6. Ex4: Divide 3x2 + 2x – 1 by x – 2 using synthetic division.
Synthetic division is a shorter method of dividing polynomials. This method can be used only when the divisor is of the form
x – c. It uses the coefficients of each term in the dividend.
Ex4: Divide 3x2 + 2x – 1 by x – 2 using synthetic division.
Since the divisor is x – 2, c = 2.
value of c
coefficients of the dividend
1. Bring down 3 2
– 1
2. (2 • 3) = 6
3. (2 + 6) = 8 6 16
4. (2 • 8) = 16 3 8 15
5. (–1 + 16) = 15
coefficients of quotient
remainder
3x + 8
Answer: 15
Synthetic Division

7. Ex5: Ex6: Use synthetic division to divide by

8. Note: Remainder Theorem
If a polynomial P (x) is divided by x –c , then the remainder equals P(c).
Ex7: Using the remainder theorem, evaluate P(x) = x 4 – 4x – 1 when x = 3.
value of x 3
– – 1 3 9 27 69 1 3 9 23 68
The remainder is 68 at x = 3, so P(3) = 68.
Note:
You can check this using substitution:
P(3) = (3)4 – 4(3) – 1 = 68.
Remainder Theorem

9. Ex8:
If P(x)=211x4-212x3 +212x2 +210x-3, then find P(1/211).
Q69/289:
Find the remainder of 5x48+6x10-5x+7 divided by x-1.
Ex9:(HW)
Find the remainder of P(x)=x103+x102+x101+x100 divided by x+i.

10. Factor Theorem
A polynomial P(x) has a factor (x – c) if and only if P(c) = 0.
Ex10: Show that (x + 2) and (x – 1) are factors of P(x) = 2x 3 + x2 – 5x + 2.
– 2 – 1 –
– 4 6
– 2 2
– 1 2
– 3 1 2
– 1
The remainders of 0 indicate that (x + 2) and (x – 1) are factors.
The complete factorization of P is (x + 2)(x – 1)(2x – 1).
Factor Theorem

11. A polynomial function of degree n has at most n real zeros.
A real number c is a zero of P (x) if and only if P(c) = 0.
Real Zeros of Polynomial Functions
If P(x) is a polynomial and c is a real number then the following statements are equivalent.
1. x = c is a zero of P.
2. x = c is a solution of the polynomial equation P (x) =0.
3. (x – c) is a factor of the polynomial P (x).
4. (c, 0) is an x-intercept of the graph of P (x).
and we have P(x)=(x-c) Q(x), where Q(x) is a polynomial of degree< degree of P(x) by 1 called the reduced polynomial
A polynomial function of degree n has at most n real zeros.
Zeros of a Function

12. Ex11: Verify that (x+4) is a factor of P(x) = 3x4+11x3-6x2-6x+8
Ex11: Verify that (x+4) is a factor of P(x) = 3x4+11x3-6x2-6x+8. and write P(x) as the product of (x+4) and the reduced polynomial Q(x).
Ex12:Prove that for any positive odd integer n, P(x) = xn +1 has x-1 as a factor.
Ex13: If x-i is a factor of the polynomial P(x) = 7x171-8x172-9x173+kx174 , then find the value of k .
The End