Precalculus: Polynomials
1. Definitions
Polynomial is made up of terms. The
monomial has one term, the binomial
two terms, and the trinomial three terms.
Examples:
3 x2 is a monomial
2 x2y + y is a binomial,
and 4 x2 - 2xy + 3 is
quadratic trinomial .
A polynomial can have no variable, one variable or more ..
Examples:
34 is a polynomial with no variable.
2 xy - 3z - 2 is a polynomial with three variables
x, y, and z.
The degree of a polynomial with one variable is the
largest exponent of that variable.
Example:
2 x5 - 3x2 + 4 x is a polynomial
of degree 5.
It is convenient to write a polynomial in its standard form,
that is the highest degree first.
Example:
a x2 + bx + c
a, b, and c are constant called coefficients.
The general form of a polynomial is written as:
anxn + ... + a2x2 + a1x1 +
a0x0
= Σakxk
k runs from 0 to n
In this chapter, we use only univariate quadratic polynomial,
that is a polynomial of the form:
a x2 + bx + c
2. Quadratic trinomials
A quadratic trinomial is a polynomial of the form:
a x2 + bx + c
Let f(x) = a x2 + bx + c
c = 0
f(x) = a x2 + bx = x(a x + b)
Example:
2 x2 - 3 x = x (2 x - 3)
b = 0
f(x) = a x2 + c = a (x2 + c/a)
if c/a <0 then f(x) = a [x + (c/a)1/2] [x - (c/a)1/2]
Example:
3 x2 + 4 = 3 x2 + 4, but
3 x2 - 4 = 3(x2 - 4/3 ) =
3(x - 2/√3 )3(x + 2/√3 )
a ≠ = 0, b ≠ = 0, c ≠ = 0
we have then to factor the whole
quadratic trinomial
a x2 + bx + c
3. Factoring quadratic trinomials
Method 1: Quadratic formula
f(x) = a x2 + bx + c = a (x - x1) (x - x2)
x1 = (- b + √Δ)/2a
x2 = (- b - √Δ)/2a
Δ = b2 - 4 a c
Example:
2 x2 - 5x + 3
Δ = b2 - 4 a c = 25 - 4 x 2 x 3 = 1
So
x1 = (- b + √Δ)/2a = (5 + 1 )/4 = 3/2
x2 = (- b - √Δ)/2a = (5 - 1 )/4 = 1
Therefore:
f(x) = 2 (x - 3/2)(x - 1)
Particular case b is even
When b is even, we can factor more rapidly as
follows:
f(x) = a x2 + 2b'x + c = a (x - x1) (x - x2)
Δ = (2b')2 - 4 a c = 4(b'2 - a c) = 4 Δ'
x1 = (- 2b' + 2 √Δ')/2a = (- b' + √Δ')/a
x2 = (- 2b' - 2 √Δ')/2a = (- b' - √Δ')/2
Example:
3 x2 - 10x + 7
Here b' = - 5. So
Δ' = (b'2 - a c) = ((- 5)2 - 3 x 7) = 4
x1 = (- 2b' + 2 √Δ')/2a = (- b' + √Δ')/a = ( 5 + 2)/3 = 7/3
x2 = (- 2b' - 2 √Δ')/2a = (- b' - √Δ')/2 = ( 5 - 2)/3 = 1
Therefore:
f(x) = 3 (x - 7/3)(x - 1)
Method 2: Completing the Square
f(x) = a x2 + bx + c = a [x2 + (b/a)x + c/a]
= a [x2 + 2(b/2a)x + c/a] =
a [x2 + 2(b/2a)x + (b/2a)2 - (b/2a)2 + c/a]
We have:
x2 + 2(b/2a)x + (b/2a)2 = (x + b/2a)2 ,
and
- (b/2a)2 + c/a = - [(b/2a)2 - c/a] =
- [(b2 - 4ac)]/ 4a2,
and
Δ = b2 - 4ac
So
f(x) = a [(x + b/2a)2 - Δ/4a2]
If Δ < 0 , we cannot factor.
If Δ > 0 , we have:
f(x) = a [(x + b/2a)2 - Δ/4a2] =
f(x) = a [x + b + Δ/2a][x + b - Δ/2a],
by using a2 - b2 = (a + b)(a - b).
Examples:
x2 + 2x + 7 = x2 + 2x + 1 - 1 + 7 =
(x + 1)2 + 6 cannot be factored.
x2 + 10x + 20 = x2 + 2(5x) + 25 - 25 + 20
= (x + 5)2 - 25 + 20 = (x + 5)2 - 5
= (x + 5)2 - (√5)2
Using a2 - b2 = (a + b)(a - b),
x2 + 10x + 20 = (x + 5 - √5)(x + 5 + √5)
Method 3: factoring by grouping
Another method for factoring quadratic trinomials is called
factoring by grouping. By some trial and error, we can find
two real m and n such as:
m n = a c
m + n = b
So
f(x) = a x2 + bx + c
= a x2 + (m + n) x + m n/a
= a x2 + m x + n x + m n/a
= x(ax + m) + n (x + m/a) = x(ax + m) + (n/a) (ax + m)
= (ax + m)(x + (n/a))
Therefore
f(x) = (1/a)(ax + m)(ax + n)
Example1:
4 x2 + 8 x + 3
m n = ac = 4 x 3 = 12
m + n = 8
Guessing gives:
m = 2 and n = 6 , so
4 x2 + 8 x + 3 = (4x + 2)(x + (6/4)) = (4x + 2)(x + 3/2)
Example2:
3 x2 - 13 x - 10
m n = ac = 3 x (- 10) = - 30
m + n = - 13
Guessing gives:
m = - 15 and n = 2 , so
3 x2 - 13 x - 10 = (3x - 15)(x + (2/3)) =
(3x - 15)(x + 2/3)
If guessing does not work, use completing the square,
or quadratic formula.
Particular case of this method
f(x) = a x2 + bx + c = a (x - x1) (x - x2)
x1 = (- b + √Δ)/2a
x2 = (- b - √Δ)/2a
Δ = b2 - 4 a c
S = x1 + x2 = (- b + √Δ)/2a + (- b - √Δ)/2a = - b/a
P = x1x2 = [(- b + √Δ)/2a ][(- b - √Δ)/2a] =
(- b)2 - Δ/(2a)2 = c/a
So
f(x) = a x2 + bx + c = a x2 + (- aS)x + aP =
= a (x2 - Sx + P)
f(x) = a x2 + bx + c = a (x2 - Sx + P)
find:
x1 + x2 = S, and
x1 x2 = P, so
f(x) = a x2 + bx + c = a (x2 - Sx + P) =
a (x - x1) (x - x2)
Example:
3 x2 - 12 x - 15 = 3 (x2 - 4 x - 45)
S = x1 + x2 = 4
P = x1 x2 = - 45
By guessing, trial and error give:
x1 = 9
x1 = - 5
4. Exercises:
Factor the following quadratic trinomials:
2x2 - 5 x + 3
3 x2 - 10 x + 7
2 x2 + 3 x - 4
x2 - 8 x + 16
x2 + 8 x + 12
x2 + x - 12
6 x2 + 7 x - 5
8 x2 + 2 x - 15
Solutions
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