Transcript Factoring by Families

Identifying “families of problems” is a great way of
approaching problems. Problems within the same
family are solved with the same process.
This PowerPoint identifies families of problems in
solving for a variable. We will be using the variable “x”,
but the variable can be any letter.
PEMDAS
 Students learn the acronym PEMDAS to help them
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remember the order of operations in solving
mathematical problems. The letters stand for:
P = Parenthesis
E = Exponents
M = Multiplication
D = Division
A = Addition
S = Subtraction
PEMDAS
 Students use the order of operation from left to right
to simplify expressions, i.e., simplifying each side of
the equation prior to working across the equality sign.
 Students use the order from right to left to solve for the
variable. Solving for the variable can also be thought
of as “peeling the onion” where the student works from
the outside to unwrap the problem and solve for the
variable.
Simplifying an expression
PEMDAS
Solving the equation for the variable
Degree 1 and constant
 If the degree of the x terms are only single order, simply
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solve for x using PEMDAS
Simplify each side by using PEMDAS from left to right
Then solve for x by using PEMDAS from right to left
Example: 2x + 2(x -4) – 3 = 11
Simplifying by distributing gives 2x + 2x -8 -3 = 11
Simplifying by gathering like terms: 4x – 11 = 11
Solving by adding 11 to each side gives 4x = 22
Dividing each side by 4 gives x = 5.5
Check by substituting x = 5.5 into the original problem
Degree 2, no Degree 1 and
constants
 If the problem has variables with only degree 2, (x2):
1. Isolate the x2 term
2. Take the square root of each side
3. Remember to include the ±
4. Example: 2x2 – 4 = 20
5. Adding 4 to each side gives: 2x2 = 24
6. Dividing each side by 2 gives: x2 = 12
7. Take square root of each side gives: x = ±2√3
8. Check by substituting ±2√3 into the original
equation
Degree ½ (Square Root), no Degree
1 and constants
 If the problem has variables with only a square root, (√x):
1. Isolate the √x term
2. Square each side
3. Solve for the variable
4. Example: 2√(x+6) – 5 = 1
5. Adding 5 to each side gives: 2√(x+6) = 6
6. Dividing each side by 2 gives: √(x+6) = 3
7. Square each side gives: x+6 = 9
8. Subtract 6 from each side gives: x = 3
9. Check by substituting 3 into the original equation
Degree 2, Degree 1 and no
constants
 If the problem has variables with degree 1 and 2, (x and
x2), but no constants:
1. Move everything to one side of the equal sign.
2. Factor out a common x.
3. Use the zero product rule to determine the solutions.
4. Example: 3x2 = 4x
5. Subtracting 4x gives: 3x2 – 4x = 0
6. Factoring out the x gives: x(3x-4) = 0
7. Either x = 0 or 3x-4 = 0, so x = 0 or x = 4/3
Four terms
 If the equation has four terms, try factoring by
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grouping and using the zero product rule:
Example: 6x2 + 18x – 8x – 24 = 0
Find the common factor of the first two and last two
terms: 6x(x + 3) – 8(x + 3) = 0
Take out the common factor of (x + 3):
(x + 3)(6x – 8) = 0
Use the zero product rule:
x + 3 = 0 or 6x – 8 = 0
x = -3 or x = 4/3
Degree 2, Degree 1, constants
and “a” = 1
 If the problem has variables with degree 1 and 2, (x and
x2) as well as constants:
1. Move everything to one side of the equal sign,
making everything equal zero, leaving the equation
in standard form of ax2 + bx + c = 0
2. If the leading coefficient (a) equals 1, find the factors
of c that add to b.
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Degree 2, Degree 1, constants
and “a” = 1
Example: x2 – x = 12
Subtracting 12 from each side gives: x2 – x – 12 = 0
The factors of -12 are: (1, -12), (2, -6), (3, -4), (4, -3), (6, 2), (12, -1)
-4 + 3 = -1, so the factors are:
(x – 4) * (x + 3) = 0
Using the zero product rule gives:
x– 4 = 0 or x + 3 = 0,
so x = 4 or x = -3
Check by substituting 4 and -3 into the original equation
Degree 2, Degree 1 and constants
and “a” ≠ 1
 If the problem has variables with degree 1 and 2, (x and
x2):
1. Move everything to one side of the equal sign,
making everything equal zero, leaving the equation
in standard form of ax2 + bx + c = 0
2. If the leading coefficient (a) does not equal 1, find
the factors of a * c that add to b.
3. Replace the “bx” term with the two factors multiplied
by x
4. Factor by grouping.
Degree 2, Degree 1 and constants
and “a” ≠ 1
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Example: 6x2 – 7x – 5 = 0
a * c = 6 * (-5) = -30, the factors of -30 are: (1, -30), (2, -15),
(3, -10), (5, -6), (6, -5), (10, -3), (15, -2) and (30, -1)
3 – 10 = -7, so substitute 3x – 10x for -7x, giving:
6x2 + 3x – 10x – 5 = 0
Factor by grouping:
3x(2x + 1) – 5(2x + 1) = 0
(2x + 1) * (3x – 5) = 0
Using the zero product rule gives
2x + 1 = 0 or 3x – 5 = 0, so
x = -1/5 or x = 5/3
Check by substituting -1/5 and 5/3 into the original
equation
If the quadratic is not factorable
 The above techniques only work if the roots (zeros,
solutions, x-intercepts) are real, and rational. They do
not work for imaginary roots, and may not work for
radical roots.
 The quadratic formula can be used to find the factors
of any quadratic equation.
 The roots can be found by:
 b  b 2  4ac
2a
 Where the a, b, and c are from the standard form
ax2 + bx + c = 0
If the quadratic is not factorable
 Example: 5x2 + 4x +4 = 0
 Using the quadratic equation:
 b  b  4ac  4  4  4 * 5 * 4  4   64
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2a
2*5
10
2
2
 So the roots are imaginary:
x = -0.4 + 0.8i or x = -0.4 – 0.8i