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A polynomial function is a function of the form
11 1 0( ) , 0n n
n n nf x a x a x a x a a
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.:
3
+ 2 2 3 1 2 xxx
Ex1: Divide x2 + 3x – 2 by x + 1 and check the answer.
x
x2 + x2x – 22x + 2
– 4
remainder
Check:
xx
xxx
22 1.
xxxx 2)1(2.
xxxxx 2)()3( 22 3.
22
2 x
xxx4.
22)1(2 xx5.
4)22()22( xx6.
(x + 2)
quotient
(x + 1)
divisor
+ (– 4)
remainder
= x2 + 3x – 2
dividend
Answer: x + 2 +1x
– 4
Dividing Polynomials
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Notes:
)(
)()(
)(
)(
xD
xRxQ
xD
xP
• 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
5210732
:Divide2
234
xxxxx
Ex2
Ex3(HW)
5230232
:2
234
xxxxx
Divide
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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.
3 2 – 12
Since the divisor is x – 2, c = 2.
3
1. Bring down 3
2. (2 • 3) = 6
6
8 15
3. (2 + 6) = 8
4. (2 • 8) = 16
5. (–1 + 16) = 15coefficients of quotient remainder
value of c coefficients of the dividend
3x + 8Answer: 2
x15
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Ex5:
Use synthetic division to divide 10112 24 xxx
by 3x
Ex6:
Use synthetic division to divide 4272 35 xxx
by 2x
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•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.
9
1 0 0 – 4 – 13
1
3
3 9
6927
23 68
The remainder is 68 at x = 3, so P(3) = 68.
You can check this using substitution: P(3) = (3)4 – 4(3) – 1 = 68.
value of x
Note:
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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.
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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.
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2 1 – 5 2– 2
2
– 4
– 3 1
– 2
0
The remainders of 0 indicate that (x + 2) and (x – 1) are factors.
– 1
2 – 3 11
2
2
– 1 0
The complete factorization of P is (x + 2)(x – 1)(2x – 1).
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A real number c is a zero of P (x) if and only if P(c) = 0.
A polynomial function of degree n has at most n real zeros.
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
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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 .