Algebraic Expressions Prepared by: Sa’diyya Hendrickson ... - Algebraic... · Factoring...

Algebraic Expressions Prepared by: Sa’diyya Hendrickson

Name: Date:

Read the following definitions and match them with the appropriate example(s) using the lines provided.

1. Variable:

A letter that is used to: (1) describe properties of a mathe-matical object, (2) represent a quantity that we dont knowor (3) be a placeholder for the input of one or more numbers.

2. Algebraic expression:

An expression that involves real numbers and at least onevariable using addition, subtraction, multiplication, divisionand exponentiation or radicals.

3. Monomial:

A real number or an expression that involves the product ofreal numbers and variables with positive integer exponents.

4. Polynomial:

A monomial or an expression that involves the additionand/or subtraction of monomials.

6. Degree of a monomial:

The sum of the exponents of the variables in a monomial.By convention, if the monomial is a real number, we saythat it has degree zero.

7. Degree of a polynomial:

The highest degree amongst the monomials in a polynomial.

8. Coefficient:

The real number in a monomial.

9. Leading coefficient of a polynomial:

The coefficient of the monomial with the highest degree.

(a) 5xy − xz3

(b) t

(c) 5x√y − 3xz

(d) From (a): 4

(e) 7

(f)4y

3 + x

(g) 8k2

(h) From (a): −1

(i) From (e): 0

(j) From (a): 5

(k) 9x4 − 2x2 + 1

(l) From (k): 9

(m)3

4

(n) From (g): k

(o) From (b): 1

(p) 2x

Making Math Possible 1 of 9 c© Sa’diyya Hendrickson

Notations Level: X

Consider the following general polynomial form:

p(x) = anxn + an−1x

n−1 + · · ·+ a2x2 + a1x+ a0

(i) The expression p(x) reads “p of x” or “p at x” and is one way torefer to a polynomial that is defined in terms of the variable x.Note that we could have replaced x with any variable of our choosing.

(ii) Because p is a polynomial, n represents some nonnegative integer (by definition).

(iii) The variables an, an−1, . . . , a2, a1, a0 represent coefficients.

Check-In

a) What does the variable n represent in the polynomial p (above)?

b) What does an represent in the polynomial p (above)?

c) Consider the polynomial p(t) = 3t5−4t2+2t−1. Determine a0, a1, a2, a3, a4 and a5.

Solution: Based on the general form above, p(t) = a5t5 + a4t

4 + a3t3 + a2t

2 + a1t+ a0.So, a0 = −1, a1 = 2 a2 = −4, a3 = 0, a4 = 0 and a5 = 3.

d) Suppose p(z) = 3z2 − 4z +1. What does p(1) mean? Determine the resulting value.

Solution: p(1) reads “p at 1” and represents the number produced by the polynomialwhen we replace z with the number 1. The resulting value is given by:

p(1) = 3(1)− 4(1) + 1

= 3− 4 + 1

= −1 + 1

= 0

e) Consider the following one-degree polynomials, p(x) = a1x+a0 and q(x) = b1x+ b0.Under which circumstances will p and q represent the same polynomial? Underwhich circumstances will they represent different polynomials?

Solution: p and q will represent the same polynomial if a1 = b1 and a0 = b0.If a1 ∕= b1 or a0 ∕= b0, then p and q will represent different polynomials.

f) Based on the information given in (i), what do you think p(x, y) might represent?If p(x, y) = y2 − 3xy, determine the value of p(−1,−2).

Solution: p(x, y) simply represents a polynomial defined in terms of two variables, xand y. Since p(x, y) = y2 − 3xy, p(−1,−2) requires that we replace x with −1 and ywith −2. This gives: p(−1,−2) = (−2)2 − 3(−1)(−2) = 4− 6 = −2.


Expanding Expressions Level: X

The most popular polynomials, defined in terms of one variable (at most), are as follows:

a) Degree = 0: constant polynomials (e.g. 5, −1, 2, etc.)

b) Degree = 1: mx+ b, where m,b ∈ R

c) Degree = 2: quadratic polynomials (e.g. ax2 + bx+ c, where a, b, c ∈ R)

d) Degree = 3: cubic polynomials (e.g. ax3 + bx2 + cx+ d, where a, b, c, d ∈ R)

** It’s all about the Distributive Property (DP)! **

Exercise E1: Expand the following expression: (x− 1)(x+ 4)

Solution: For the given expression:

a! "# $(x− 1)(

b!"#$x +

c!"#$4 ), we can let a = x+ 1,

b = x and c = 4. This results in the following expansion:

Notice that each term in the first pair of brackets was simply multiplied witheach term in the 2nd pair of brackets! In general, we have:

Many people are familiar with the acronym FOIL (First, Outer, Inner, Last).But this is just a special case of the Distributive Property, when we have exactlytwo elements in each pair of brackets!



The Distributive Property tells us to simply take each term inthe 1st pair of brackets (one at a time) and multiply them with eachterm in the 2nd pair of brackets (i.e. distribute them). Unlike FOIL,this works for any number of elements in each pair of brackets!

For the above exercise, our strategy gives:

(x− 1)(x+ 4)DP= (x)(x) + (4)(x)− (1)(x)− (1)(4) as before, but in one step!

= x2 + 4x− x! "# $like terms

−4

collect= x2 + 3x− 4

Exercise E2: Expand (a− b)3

Solution: First recall that (a−b)3 = (a−b)(a−b)(a−b), by def ’n of exponenti-ation for a positive integer exponent. So, our approach is to use the associativeproperty of multiplication to group and expand two at a time!

(a− b)3 = (a− b)(a− b)! "# $(a− b)

DP= (a2 − ab− ba+ b2)(a− b) Notice that ab = ba

When expanding, it’s a good idea to write the letters in alphabeticalorder because it makes it easier to identify like terms !!

collect= (a2 − 2ab+ b2)(a− b)

DP= (a2)(a) + (a2)(−b) −(2ab)(a)− (2ab)(−b) +(b2)(a) + (b2)(−b)

= a3 − a2b− 2a2b+ 2ab2 + ab2 − b3

collect= a3 − 3a2b+ 3ab2 − b3

Try the following exercises on your own, using the same strategy:

1) Prove that (a+ b)3 = a3 + 3a2b+ 3ab2 + b3.2) Prove that (a− b)(a2 + ab+ b2) = a3 − b3.



Exercise E3: Expand the expression −2ab2c (5a3 − b+ 3b2c)

As you distribute each term, it may be helpful to: first record the productof the constants, while carefully determining the sign (i.e. is the product positive ornegative?). Then, determine and record the product of each variable, one by one.(The chart below describes how to organize your mental math, as you expand.)

−2ab2c (5a3 − b+ 3b2c)

Distribution Constants a b c Summary

First (−2)(5) = −10 a · a3 = a4 b2 c −10a4b2c

Second (−2)(−1) = 2 a

Third b2 · b2 = b4 c · c = c2

Solution:

−2ab2c (5a3 − b+ 3b2c)DP=

P.O.E− 10a4b2c+ 2ab3c− 6ab4c2

Exercise E4: Expand the expression 4(x2 − 3x+ 5)− 3(x2 − 2x+ 1)

Solution:

4(x2 − 3x+ 5)− 3(x2 − 2x+ 1)DP= 4x2 − 12x+ 20− 3x2 + 6x− 3

collect= 4x2 − 3x2−12x+ 6x+20− 3

collect= x2 − 6x+ 17

Exercise E5: Expand and evaluate (2−√3)(2 +

√3)

Solution:

(2−√3)(2 +

√3)

DP= 2(2) + 2

√3− 2

√3− (

√3)(

√3)

= 4− (√3)2

= 4− 3 since ( n√a)n = a, for a ≥ 0

= 1


Factoring Expressions Level: X

** It’s still all about the Distributive Property (DP)! **

On the previous pages, we were going from factored form to expanded form.Now, we will begin with expanded form and end in factored form .We’ll know that we’ve achieved factored form once the original expression is aproduct of terms and/or smaller expressions.

Exercise F1: Factor the expression −12x4y2 + 30xy3 − 18xy4z.

S1 Decide if you’d like to factor out “−1” from all terms. This isrecommended if your leading term is negative.

Because our leading term is −12x4y2, which is negative, we will pull−1 as a common factor. Note that we can always do this!

e.g. −(12x4y2−30xy3+18xy4z)

S2 Determine the GCF (Greatest Common Factor) among theconstants and each variable, one at a time. Note that for each variable,we must factor the lowest power among all terms. Otherwise, wecannot guarantee that we have a common factor.

Expression: − 12x4y2 + 30xy3 − 18xy4z

Elements GCF

Factor−1? N/A Yes/No: −1

Constants 12, 30, 18 2 · 3 = 6

x x4, x1, x1 x1

y y2, y3, y4

z z0, z0, z1 z0 = 1

Summary N/A −6xy2

This chart gives us a visual representation of how we can organize ourapproach if we were to complete this strategy using mental math.



Exercise F1 (Continued)

S3 Factor out the GCF of each element, starting with the constantsand then each variable. Then complete the expression in the bracketsby writing in the missing part of the original term.

−12x4y2 + 30xy3 − 18xy4z = − (12x4y2−30xy3+18xy4z)

= − 6 ( 2!"#$12÷6

x4y2 − 5!"#$30÷6

xy3 + xy4z)

P.O.E= − 6x1 (2 x4−1

! "# $x4 ÷ x1

y2 − 5 y3 + 3 y4z)

P.O.E= − 6xy2 (2x3 y2−2 − 5 + 3 z)

= − 6xy2 (2x3 − 5y + 3y2z)

With practice, you will be able to do this in one line!

Let’s begin by recalling the following definition:

Perfect square: a perfect square is an expression of form a2, wherea is any integer. Note that there is nothing specific about our choiceof the letter a. Any letter will do! e.g. b2, c2, s2, t2, x2, y2, etc.

Below are some square numbers that appear quite often. Therefore, itmay be worthwhile to commit them to memory.

02 = 0 12 = 1 22 = 4 32 = 9 42 = 16 52 = 25

62 = 36 72 = 49 82 = 64 92 = 81 102 = 100

I. Difference of Squares: a2 − b2 = (a + b)(a − b)

proof : (a− b)(a+ b)DP= a2 + ab− ab! "# $

= 0

−b2

collect= a2 − b2 □

Exercise F2: Factor the expression x2 − 1.

S1 Determine if there are exactly two perfect squares, seper-ated by a “−” sign. Then, express the terms as powers of 2.

x2 "1 = 12 "



Exercise F2 (Continued)

S2 Identify “a” and “b” based on the given identity.

x2 − 1 = x2 − 12 ⇒ a = x and b = 1

S3 Substitute into the identity to obtain the factored form.

x2 − 1 = x2 − 12 a = x and b = 1

= (x a 1)(x a 1)

= (x+ 1)(x− 1)

Exercise F3: Factor the following expressions.

a) 9− z2

b) 4k2 − 49

c) 16u2 − 25v2

a) Solution : 9− z2 = 32 − z2 a = 3 and b = z

= (3 a z)(3 a z)

= (3 + z)(3− z)

b) Solution : 4k2 − 49 = 22k2 − 72

P.O.E= (2k)2 − 72 a = 2k and b = 7

= (2k a 7)(2k a 7)

= (2k + 7)(2k − 7)

c) Solution : 16u2 − 25v2 = 42u2 − 52v2

P.O.E= (4u)2 − (5v)2 a = 4u and b = 5v

= (4u a 5v)(4u a 5v)

= (4u+ 5v)(4u− 5v)



II. Perfect Square: (a ± b)2 = a2 ± 2ab+ b2

proof : (a ± b)2 = (a ± b)(a ± b)

DP= a2 ± ab± ab+ (±b)2

collect= a2 ± 2ab+ b2 □

Exercise F4: Factor the expression 4x2 − 12x+ 9.

S1 Determine if the expression is a trinomial with perfectsquares as its first and last terms. Then, express the first andlast terms as powers of 2.

4x2 = 22x2 P.O.E= (2x)2 "

9 = 32 "

S2 Identify what “a” and “b” may be based on the identityand info from S1. Note: we still have to verify the middle term!

4x2 − 12x+ 9 = (2x)2 − 12x+ (3)2 ⇒ a = 2x and b = 3

S3 Verify that the middle term equals ±2ab.

For a = 2x and b = 3 ⇒ −2ab = −2(2x)(3) = −12x "

S4 Substitute into the identity to obtain the factored form.

4x2 − 12x+ 9 = (2x)2 − 2(3)(2x) + (3)2 a = 2x and b = 3

= (2x a 3)2 by identity

= (2x− 3)2 since the middle term is negative

Exercise F5: Factor 9s2t2 − 6st+ 1.

Solution : 9s2t2 − 6st+ 1P.O.E= (3st)2 − 2(3)st+ (1)2 a = 3st and b = 1

= (3st a 1)2 by identity

= (3st− 1)2


Algebraic Expressions Prepared by: Sa’diyya Hendrickson ... - Algebraic... · Factoring...

Documents

Transcript of Algebraic Expressions Prepared by: Sa’diyya Hendrickson ... - Algebraic... · Factoring...