5 6 substitution and factoring formulas

Substitution and Factoring Formulas
Frank Ma © 2011

We evaluated expressions with numerical inputs.

For example, evaluate A2 – B2 with A = 2 and B = 1 as inputs

For example, evaluate A2 – B2 with A = 2 and B = 1 as inputs, we have 3.

For example, evaluate A2 – B2 with A = 2 and B = 1 as inputs, we have 3. We extend this to evaluating with expressions as inputs.

For example, evaluate A2 – B2 with A = 2 and B = 1 as inputs, we have 3. We extend this to evaluation with expressions as inputs. We use the word “substitute” for this type of procedures.

Example A. Evaluate A2 – B2 with A = x + 2 and B = 2x. Expand and simplify the result.

Replace A by (x + 2), and B by (2x)

We also say that we
“plug in” (x + 2) for A

make sure to encase the inputs in the ( )’s
We also say that we

we have (x + 2)2 – (2x)2
We also say that we

we have (x + 2)2 – (2x)2
We also say that we
= x2 + 4x + 4– 4x2
= –3x2 + 4x + 4

we have (x + 2)2 – (2x)2
We also say that we
= x2 + 4x + 4– 4x2
= –3x2 + 4x + 4
Reverse the procedure.

we have (x + 2)2 – (2x)2
We also say that we
= x2 + 4x + 4– 4x2
= –3x2 + 4x + 4
Reverse the procedure. Sometime the inputs can be easily identified when the formula and the outcome are given.

we have (x + 2)2 – (2x)2
We also say that we
= x2 + 4x + 4– 4x2
= –3x2 + 4x + 4
Example B. Evaluate A2 – B2 and we obtain
a. 4x2– 9y2, what are A and B?

we have (x + 2)2 – (2x)2
We also say that we
= x2 + 4x + 4– 4x2
= –3x2 + 4x + 4
Match the A2 = 4x2, we’re asking ( ? )2 = 4x2,

we have (x + 2)2 – (2x)2
We also say that we
= x2 + 4x + 4– 4x2
= –3x2 + 4x + 4
Match the A2 = 4x2, we’re asking ( ? )2 = 4x2, so A = (2x).

we have (x + 2)2 – (2x)2
We also say that we
= x2 + 4x + 4– 4x2
= –3x2 + 4x + 4
Match the A2 = 4x2, we’re asking ( ? )2 = 4x2, so A = (2x).
Similarly we conclude that B = (3y).

b. 25x2y4– 1, what are A and B?

Match the A2 = 25x2y4 and B2 = 1,

Match the A2 = 25x2y4 and B2 = 1, we have that
A = 5xy2 and B = 1.

A = 5xy2 and B = 1.
c. x2+ 2xy + y2 – 9, what are A and B?

A = 5xy2 and B = 1.
Factorx2 + 2xy + y2

A = 5xy2 and B = 1.
Factorx2 + 2xy + y2we have (x + y)(x + y) = (x + y)2,

A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.

A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
Hence A = (x + y) and B = 3.

A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
Example C. Evaluate A3 – B3with inputs A and B
a. we have x3 – 125, what are A and B?

A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
Match the A3 = x3, we have A = x.

A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
Similarly we’ve B3 = 125 so B = 5.

A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
b. we have 8y3 – (2 – y)3 , what are A and B?

A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
Match the A3 =8y3, we have A = 2y.

A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
By inspection, we have B = (2 – y).

A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
By inspection, we have B = (2 – y).
Recall the conjugate product formula
(A – B)(A + B) = A2 – B2.

The reverse of the conjugate product formula is the
factoring formula for difference of squares :
A2 – B2 = (A – B) (A + B)

A2 – B2 = (A – B) (A + B)
As suggested by the name, we use this formula when
it is the subtraction of two perfect square terms.

A2 – B2 = (A – B) (A + B)
Example D.
a. x2 – 25

A2 – B2 = (A – B) (A + B)
Example D.
a. x2 – 25
= (x)2 – (5)2

A2 – B2 = (A – B) (A + B)
Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)

A2 – B2 = (A – B) (A + B)
Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
b. 4x2 – 9y2

A2 – B2 = (A – B) (A + B)
Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
b. 4x2 – 9y2
= (2x)2 – (3y)2

A2 – B2 = (A – B) (A + B)
Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
b. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)

A2 – B2 = (A – B) (A + B)
Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
b. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
c. 25A2B4 – 1

A2 – B2 = (A – B) (A + B)
Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
b. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
c. 25A2B4 – 1
= (5AB2)2 – (1)2

A2 – B2 = (A – B) (A + B)
Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
b. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
c. 25A2B4 – 1
= (5AB2)2 – (1)2
= (5AB2 – 1)(5AB2 + 1)

d. (A + 2B)2 – 16

d. (A + 2B)2 – 16
=(A + 2B)2 – (4)2

d. (A + 2B)2 – 16
=(A + 2B)2 – (4)2
= ((A + 2B)– 4)((A + 2B)+ 4)

d. (A + 2B)2 – 16
=(A + 2B)2 – (4)2
= ((A + 2B)– 4)((A + 2B)+ 4)
= (A + 2B – 4)(A + 2B + 4)

d. (A + 2B)2 – 16
=(A + 2B)2 – (4)2
= ((A + 2B)– 4)((A + 2B)+ 4)
= (A + 2B – 4)(A + 2B + 4)
Caution: The expressions
A2 + B2 or – A2 – B2
are not factorable!

d. (A + 2B)2 – 16
=(A + 2B)2 – (4)2
= ((A + 2B)– 4)((A + 2B)+ 4)
= (A + 2B – 4)(A + 2B + 4)
A2 + B2 or – A2 – B2
are not factorable!
Sum and Difference of Cubes

d. (A + 2B)2 – 16
=(A + 2B)2 – (4)2
= ((A + 2B)– 4)((A + 2B)+ 4)
= (A + 2B – 4)(A + 2B + 4)
A2 + B2 or – A2 – B2
are not factorable!
Both the difference and the sum of cubes may be factored.

d. (A + 2B)2 – 16
=(A + 2B)2 – (4)2
= ((A + 2B)– 4)((A + 2B)+ 4)
= (A + 2B – 4)(A + 2B + 4)
A2 + B2 or – A2 – B2
are not factorable!
If we try to factor A3 – B3, we might guess that
A3 – B3 = (A – B)(A2 + B2)

d. (A + 2B)2 – 16
=(A + 2B)2 – (4)2
= ((A + 2B)– 4)((A + 2B)+ 4)
= (A + 2B – 4)(A + 2B + 4)
A2 + B2 or – A2 – B2
are not factorable!
If we try to factor A3 – B3, we might guess that
A3 – B3 = (A – B)(A2 + B2)
But this does not work because we have the inner
and outer products –BA2 and AB2remaining in the expansion as shown above.

But if we put in the adjustment +AB, then all the cross-
products with A and B are eliminated.

products with A and B are eliminated. That is
(A – B)(A2+ AB + B2)

(A – B)(A2+ AB + B2)
= A3 + A2B + AB2 – BA2 – B2A – B3

(A – B)(A2+ AB + B2)
= A3 + A2B + AB2 – BA2 – B2A – B3
= A3 – B3

(A – B)(A2+ AB + B2)
= A3 + A2B + AB2 – BA2 – B2A – B3
= A3 – B3
Hence A3 – B3 = (A – B)(A2 + AB + B2)

(A – B)(A2+ AB + B2)
= A3 + A2B + AB2 – BA2 – B2A – B3
= A3 – B3
Hence A3 – B3 = (A – B)(A2 + AB + B2)
Through a similar argument we have that
A3 + B3 = (A + B)(A2 – AB + B2)

(A – B)(A2+ AB + B2)
= A3 + A2B + AB2 – BA2 – B2A – B3
= A3 – B3
Hence A3 – B3 = (A – B)(A2 + AB + B2)
A3 + B3 = (A + B)(A2 – AB + B2)
We write these two formulas together as
A3 B3 = (A B)(A2 AB + B2)
–
+
+
–
–
+

(A – B)(A2+ AB + B2)
= A3 + A2B + AB2 – BA2 – B2A – B3
= A3 – B3
Hence A3 – B3 = (A – B)(A2 + AB + B2)
A3 + B3 = (A + B)(A2 – AB + B2)
We write these two formulas together as
A3 B3 = (A B)(A2 AB + B2)
–
+
+
–
–
+
Note that the signs are reversed for the adjustments in
these formulas.

Example C.
a. 8x3 – 27

Example C.
a. 8x3 – 27(Note that both 8 and 27 are cubes)

Example E.
= (2x)3 – (3)3

Example E.
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)

Example E.
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)

Example E.
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125

Example E.
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125 (both 64 and 125 are cubes)

Example E.
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
= (4A)3 + (5)3

Example E.
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)

Example E.
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)

Example E.
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
Summary on Factoring

Example E.
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
I. Always factor out the GCF first.

Example E.
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
II. Make sure the leading term is positive and all terms are
arranged in order.

Example E.
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
arranged in order.
III. If left with a trinomial, use reverse-FOIL or ac-method.

Example E.
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
arranged in order.
IV. If it is a two-term expression, try factoring by formula.

Example E.
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
arranged in order.
IV. If it is a two-term expression, try factoring by formula.
V. If it is a four term expression, try the grouping-method.

Example F. Factor completely.
A. 20A3 – 45A

A. 20A3 – 45A Take out the common factor
= 5A(4A2 – 9)

= 5A(4A2 – 9)Difference of square
= 5A(2A – 3)(2A + 3)

= 5A(2A – 3)(2A + 3)
B. –20A3 + 45A2 – 10A

= 5A(2A – 3)(2A + 3)
B. –20A3 + 45A2 – 10A Take out the common factor and -1
= –5A(4A2 – 9A + 2)

5 6 substitution and factoring formulas

by math123a on Jun 22, 2011

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5 6 substitution and factoring formulas — Presentation Transcript