Transcript x 2 + bx + c

MTH 065 – Elementary Algebra
Chapter 6
Review
The Graph of f(x) = x2 + bx + c
Solutions of
vs.
x2
+ bx + c = 0
vs.
Factorization of x2 + bx + c

y
The graph is a parabola.

x





The solutions are the x-intercepts.

The factors are (x – m)(x – n),
where m and n are the x-intercepts.
The Graph of f(x) = x2 + bx + c
Solutions of
vs.
x2
+ bx + c = 0
vs.
Factorization of x2 + bx + c
Three possibilities …
x = 3, –1
(x – 3)(x + 1)
x = –2
(x + 2)2
No Solution
Does not Factor
What if an x-intercept is the origin?
The Principle of Zero Products
 If ab = 0, then either a = 0, b = 0, or both.
 Using this Principle
 Solve:
x(x – 5)(2x + 1) = 0
 Solution:
x=0
x–5=0  x=5
2x + 1 = 0  x = –½
 Therefore …
x = 0, 5, –½
Factoring – Step 1 … always!
 Factor out common factors.
 Examples:
 Constant:
6x2 – 12x – 21 = 3(2x2 – 4x – 7)
 Variable:
x3 + 5x2 – 2x = x(x2 + 5x – 2)
 Both:
–4x4 + 32x3 – 20x2 = –4x2(x2 – 8x + 5)
Factoring by Grouping (4 terms)
x3 – 2x2 + 5x – 10
 Make two groups of two terms.
(x3 – 2x2) + (5x – 10)
 Factor out common factors from each group.
x2(x – 2) + 5(x – 2)
 If the remaining binomials are identical, factor them out.
(x – 2)(x2 + 5)
NOTE: Be careful with the sign of the third term when it is negative!
Factoring:
2
x
+ bx + c
 Leading coefficient = 1
 Find two numbers m & n where …
mn = c
m+n=b
 Then the factors are …
(x + m)(x + n)
 Signs …
 If c is positive, then both numbers have the same sign as b.
 If c is negative, then the “larger” one has the same sign as b and
the other one has the opposite sign.
Solving:
2
x
+ bx + c = 0
 All terms to the left or right (one side MUST be 0).
 Leading coefficient = 1
 Factor the polynomial …
(x + m)(x + n) = 0
 Set each factor equal to 0 and solve.
 i.e. Solutions are:
x = – m, –n
Factoring:
2
ax
+ bx + c
 Leading coefficient > 1
 Find two numbers m & n where …
mn = ac
m+n=b
 Rewrite the polynomial: ax2 + mx + nx + c
 Factor by grouping (be careful when nx is negative).
 Signs …
 If ac is positive, then both numbers have the same sign as b.
 If ac is negative, then the “larger” one has the same sign as b and
the other one has the opposite sign.
Solving:
2
ax
+ bx + c = 0
 All terms to the left or right (one side MUST be 0).
 Leading coefficient > 1
 Factor the polynomial (ac method) …
(rx + s)(tx + u) = 0
 Set each factor equal to 0 and solve.
 i.e. Solutions are:
x = – s/r, –u/t
Leading Coefficient Negative
 Factor out –1 first, and then proceed as before.
 Example:
–8x2 + 10x + 3
= –[8x2 – 10x – 3]
= –[8x2 – 12x + 2x – 3]
= –[(8x2 – 12x) + (2x – 3)]
= –[4x(2x – 3) + 1(2x – 3)]
= –(2x – 3)(4x + 1)
Factoring: Special Case #1
 Perfect Square Polynomials
ax2 + bx + c
 If a is a perfect square … a = m2
 If c is a perfect square … c = n2
 If b = 2mn
 Then it factors as …
 If b is positive: (mx + n)2
 If b is negative: (mx – n)2
Factoring: Special Case #2
 Difference of two squares.
ax2 – c
 If a is a perfect square … a = m2
 If c is a perfect square … c = n2
 Then it factors as …
(mx + n)(mx – n)
NOTE: It must be subtraction! If it is addition, it will not factor.
Factoring Polynomials in Quadratic Form
af(x) 2 + bf(x) + c
 Let y = f(x)
(i.e. replace f(x) with y)
ay2 + by + c
 Factor …
(my + n)(ry + s)
 Substitute back (i.e. replace y with f(x)
(mf(x) + n)(rf(x) + s)
 Simplify … if possible.
Applications …
1.
Familiarize - read & summarize
1.5
Estimate - approximation
2.
Translate - equation
3.
Carry out - solve & answer
4.
Check - reasonable
5.
State - answer w/ units