Download FTCE Middle Grades Math 5-9 Skills 6.13

FTCE Middle Grades Math 5-9
Skills 6.13-6.25
Factoring
Example: Factor the trinomial:
Factor out GCF:
1.
2.
3.
Find GCF of coefficients.
Find smallest power of each variable.
Multiply the terms from steps 1 and 2, this is
the GCF.
Divide each term in the polynomial by the
GCF.
4.
Example: Factor the polynomial:
2 x 2 y 3 z 4 + 4 xyz 3 + 10 x 4 y 2 z 2 =
2 xyz ( xy z + 2 z + 5 x y )
2
2 2
2
No GCF. Multiply 2 x 6 = 12. What two factors of
12 add up to the coefficient of the middle term? 4
and -3. Split the middle:
2 x 2 + 4 x − 3x − 6
(Notice the polynomial has not changed)
Factor by grouping:
2 x 2 + 4 x − 3x − 6 =
2 x ( x + 2 ) − 3( x + 2) =
( 2 x − 3)( x + 2 )
Factor by Grouping
Special Factoring Patterns:
Factoring by grouping is usually done when you have
4 terms and they do not look to have anything in
common.
Difference of Squares: a − b
Difference of Cubes (SOAP):
1.
2.
3.
4.
Remove the GCF if possible.
Group terms with common factors.
Remove the GCF of each group.
Use the distributive law to rewrite the
expression.
Example: Factor
6 x2 − 9 x + 4 x − 6
6 x2 − 9 x + 4 x − 6 =
3 x( 2 x − 3) + 2( 2 x − 3) =
(3x + 2)( 2 x − 3)
Factoring Trinomials with a Leading Coefficient
of 1:
1.
2.
Factor out the GCF if possible.
Take factors of 3rd term that add up to the
middle term and factor into 2 binomials.
Example: Factor the trinomial:
x2 + x − 6 =
( x + 3)( x − 2)
Factoring Trinomials with a Leading Coefficient
other than 1:
1.
2.
3.
4.
5.
Factor out the GCF if possible.
Multiply leading coefficient and 3rd term.
Find the factors that add to the coefficient of
the middle term.
Split the middle term.
Factor by grouping.
2 x2 + x − 6
2
2
= (a + b)(a − b)
a3 − b3 = ( a − b)( a 2 + ab + b2 )
Sum of Cubes (SOAP):
a3 + b3 = (a + b)(a 2 − ab + b2 )
SOAP: Same, Opposite, Always Positive
Solving Quadratic Equations:
ax 2 + bx + c = 0
If the quadratic is factorable, factor, set each of the
factors equal to 0 and solve.
If the quadratic is not factorable, use the quadratic
formula:
x=
−b ± b2 − 4ac
2a
Solving Quadratics by Completing the Square:
Used to rearrange the quadratic into a perfect square
so it can be solved.
1.
2.
3.
4.
5.
6.
Move the constant term to the right side.
Divide through by whatever is multiplied on
the squared term on both sides of the
equation.
Take half of the coefficient of the x-term,
and square it. Add this square to both sides
of the equation.
Convert the left-hand side to squared form,
and simplify the right-hand side.
Take square root of both sides.
Solve for x.
FTCE Middle Grades Math 5-9
Example: Solve by completing the square:
Skills 6.13-6.25
Graphing Quadratic Inequalities
2 x − 12 x − 16 = 0
2
2 x − 12 x
= 16
x − 6x
=8
2
2
3.
x − 6 x + ( −3) = 8 + ( −3)
2
1.
2.
2
2
x2 − 6x + 9 = 8 + 9
4.
( x − 3)2 = 17
( x − 3)2 = 17
Graph as if it is a quadratic equation.
Use a dotted line for > or < symbols. Use a
solid curve for ≥ or ≤ symbols.
Select a point inside the parabola and
substitute the x and y values of that point
into the function.
If the point is a solution of the inequality,
shade inside the parabola. If the point is not
a solution, shade outside of the parabola.
Solving Radical Equations (Square Roots Only)
( x − 3) = ± 17
1.
2.
3.
4.
x = 3 ± 17
Applications of the Discriminant
Isolate the variable.
Square both sides.
Solve for x.
Check solution(s).
The discriminant is the name given to the expression
that appears under the square root (radical) sign in the
quadratic formula.
Example: Solve
b 2 − 4ac
(
The discriminant tells you about the "nature" of the
roots of a quadratic equation given that a, b and c are
rational numbers. It tells you the number of real
roots associated with a quadratic equation.
x − 5 = 100
x = 105
If
If
If
b − 4ac is positive, there are two real roots.
b 2 − 4ac is zero, there is one real root.
b 2 − 4ac is negative, there are no real roots.
x − 5 = 10 .
x − 5 = 10
x −5
)
2
= 102
Domain and Range
2
Graphing Quadratic Functions
y = ax 2 + bx + c
If a is positive, the parabola opens up.
If a is negative, the parabola opens down.
To graph a parabola:
1. Find the x-coordinate of the vertex using the
formula x =
2.
3.
4.
−b
.
2a
To find the y-coordinate of the vertex,
substitute the x-coordinate of the vertex into
the function.
To find additional points on the parabola,
pick x-values on both sides of the vertex and
substitute into the function to find the
corresponding y-values.
Plot the points found in step 3 and connect
with a smooth curve.
A function assigns a unique value to each input
value. For each x, there is only one value of y.
The domain of a function is the set of all possible x
values which will make the function "work" and will
output real y-values.
When finding the domain, remember:
• The denominator of a fraction cannot be
zero.
• The values under a square root sign must be
positive.
Direct and Inverse Variation
Direct Variation
y = kx
“y varies as x”
Inverse Variation
y=
“y varies inversely as x”
k
x
Solving Linear Systems of Equations by Graphing
Graph both equations on one set of axes. The point
of intersection is the solution of the system.
FTCE Middle Grades Math 5-9
Solving Linear Systems of Equations by
Substitution
1.
2.
3.
Solve one of the equations for one of the
variables.
Substitute what you found for the chosen
variable back into the other equation and
solve for the other variable.
Solve for the first variable.
Example: Solve the following linear system of
equations using the substitution method.
Skills 6.13-6.25
Example: Solve the following linear system of
equations using the elimination method.
4 x + 2 y = −4
−x + y = 4
1.
4 x + 2 y = !4
!4 x + 4 y = 16
2.
2.
Solve for y in the second equation to
get y = x + 4 .
Substitute into first equation for y and solve
for x.
4 x + 2( x + 4) = −4
4 x + 2 x + 8 = −4
6 x = −12
3.
x = −2
Solve for y by substituting the value found
for x.
4 x + 2 y = −4
4( −2) + 2 y = −4
−8 + 2 y = − 4
2y = 4
y=2
The solution is (-2, 2).
Solving Linear Systems of Equations by
Elimination
1.
2.
3.
Multiply or divide either or both of the
equations by a number so that the
coefficients of one variable are opposites of
each other.
Add the equations to eliminate one of the
variables.
Substitute what you found for the chosen
variable back into the other equation and
solve for the eliminated variable.
Add the equations together.
4 x + 2 y = −4
−4 x + 4 y = 16
6 y = 12
y=2
4 x + 2 y = −4
−x + y = 4
1.
Multiply the second equation by 4 so that
the coefficients of the x terms are opposites.
3.
Solve for x by substituting the value found
for y into one of the original equations.
4 x + 2 y = −4
4 x + 2( 2) = −4
4 x + 4 = −4
4 x = −8
x = −2
The solution is (-2, 2).
Properties of Real Numbers:
For any set of numbers to be closed under an
operation, the result of that operation must be
included within that set of numbers.
Closure for Whole-Number Addition – the sum of
any two whole numbers is a whole number.
Closure for Whole-Number Mult. – the product of
any two whole numbers is a whole number.
Commutative Property
Addition: a + b = b + a Multiplication: ab = ba
Associative Property
Addition: a + (b + c ) = ( a + b) + c
Multiplication:
a(bc) = ( ab)c
Distributive Property
a(b + c ) = ab + ac
a(b − c ) = ab − ac
Additive Identity: a + 0 = a
Multiplicative Identity: a g1 = a
a + −a = 0
1
Multiplicative Inverse: a g = 1
a
Additive Inverse: