Special Products And Factoring

Question 1

the result when we multiply polynomials in some special cases which include:

Multiplying a binomial by another binomial (FOIL method)Squaring a binomial (multiplying a binomial by itself)Difference of two squares (multiplying binomials with the same terms but with opposite signs)Cubing a binomial (multiplying a binomial to itself thrice)

Answer 1

Special products

Question 2

FOIL stands for

Answer 2

First Terms,OuterTerms,Inner terms, andLast terms. The FOIL method is a technique used to multiply two binomials

Question 3

How to use the FOIL method in 5 steps.

Answer 3

Multiply the first terms of the binomials.Multiply the outer terms of the binomials.Multiply the inner terms of the binomials.Multiply the last terms of the binomials.Combine like terms.

Question 4

Example 1:Compute for (x + 3)(x + 2).

Answer 4

Solution:Let us use the FOIL method to compute for (x + 3)(x + 2).

Step 1: Multiply the first terms of the binomials.The first terms of the binomials are both x. Hence, x times x is equal to x2.

Step 2: Multiply the outer terms of the binomials. The outer terms are the first term of the first binomial and the last term of the second binomial. That is, x times 2 is equal to 2x.

Step 3: Multiply the inner terms of the binomials.The inner terms are the second term of the first binomial and the first term of the second binomial. That is, 3 times x is equal to 3x.

Step 4: Multiply the last terms of the binomials. The last terms of (x + 3) and (x + 2) are 3 and 2 respectively. Thus, 3 x 2 = 6

Step 5: Combine like terms.So far, we have obtained x2+ 2x + 3x + 6. Note that we can combine 2x and 3x since they arelike terms.

Therefore, using the FOIL method,
(x + 3)(x + 2) =x^2+ 5x + 6

Question 5

Example 2:Use the FOIL method to multiply
(3x – 1) by (x + 7)

Answer 5

Solution:

Step 1: Multiply the first terms of the binomials

Step 2: Multiply the outer terms of the binomials

Step 3: Multiply the inner terms of the binomials

Step 4: Multiply the last terms of the binomials

Step 5: Combine like terms

Hence, using the FOIL method,
(3x – 1)(x + 7) =3x^2+ 20x – 7.

Question 6

Example 1:Expand (x + 3) ^2

Answer 6

Solution:

Step 1: Square the first term of the binomial. The first term of the binomial is x. Squaring x means raising it to the power of 2. Therefore, the square of x is simply x2.

Step 2: Multiply the product of the first and second term of the binomial by 2. The first term is x while the second term is 3. Multiplying them together, we have 3 times x or 3x. Afterward, we multiply the result by 2. Therefore, we have 6x.

Step 3: Square the last term of the binomial. The last term of the binomial is 3 and its square is equal to 32= 9.

Step 4: Combine the results you have obtained from Step 1 to Step 3.Combining the results we have obtained from the first three steps, we have x2+ 6x + 9

Therefore, (x + 3)^2=x^2+ 6x + 9

Question 7

Example 2:Expand (p – q) ^2

Answer 7

Thus, (p – q)^2=p^2– 2pq + q^2

Question 8

Example 3:Compute for (2z – 3)(2z – 3)

Answer 8

Solution:We can express (2z – 3)(2z – 3) as (2z – 3)2. This means that we can apply the steps on squaring a binomial to determine the answer to (2z – 3)(2z – 3).

Thus, (2z – 3)(2z – 3) =4z^2– 12z + 9

Question 9

Example 4:What is the product of (5x – 2)(5x – 2)?

Answer 9

Solution:We can express (5x – 2)(5x – 2) as (5x – 2)2.This means that we can apply the steps on squaring a binomial to determine the answer to (5x – 2)(5x – 2).

Therefore, (5x – 2)(5x – 2) =25x^2– 20x + 4

Question 10

For instance, let us try to expand (2w – 3)^2mentally.

Answer 10

The square of the first term (i.e., 2w) is 4w2.

Multiply the first and second terms: 2w times – 3 is -6w. Multiply -6w by 2, we have -12w.

Then, we square the last term which is -3: (-3)2= 9

Combining what we have obtained, we have
4w^2– 12w + 9.

Question 11

Example 1:Compute for (x + 3)(x – 3).

Answer 11

Solution:Since the binomials have the same terms but with opposite signs, we can conclude that the result will be a difference of two squares.

To obtain the answer, we just square the first term (which is x) to obtain x2and also square the second term (which is 3) so we obtain 9.

Now, since we have concluded earlier that the result is a difference of two squares, we just put a minus sign between x2and 9.

Therefore, (x + 3)(x – 3) =x^2– 9

Question 12

Example 2:What is the product of (5y – a)(5y + a)?

Answer 12

Solution:Since the binomials have the same terms but with opposite signs, we can conclude that the result will be a difference of two squares.

Squaring the first term: (5y)2= 25y2

Squaring the second term: (a)2= a2

Thus, the answer is25y^2– a^2

Question 13

Example 3:Compute for (a^2+ b^2)(a^2– b^2)

Answer 13

Solution:Since the binomials have the same terms but with opposite signs, we can conclude that the result will be a difference of two squares.

Squaring the first term: (a2)2= a4(take note that we apply thepower rulehere)

Squaring the second term: (b2)2= b4

Therefore, the answer isa^4– b^4

Question 14

Example 4:Multiply (1 – 3p)(1 + 3p)

Answer 14

Solution:

Squaring the first term: (1)2= 1

Squaring the second term: (-3p)2= 9p2

Therefore, (1 – 3p)(1 + 3p) =1 – 9p^2

Question 15

Example 5:What is [(x + y) – 2][(x + y) + 2] in expanded form?

Answer 15

Solution:Although it seems that there are three terms involved in each expression, we can consider(x + y)as a single term in this case since it is grouped using a parenthesis. Thus, we have two terms for each expression and can consider them binomials.

Since the binomials have the same terms but with opposite signs, we can conclude that the result will be a difference of two squares.

Squaring the first term: (x + y)2

Squaring the second term: (2)2= 4

Thus, we have(x + y)^2– 4

However, note that (x + y)2can be expanded further since it is a square of a binomial.

Applying the steps on squaring a binomial. We have: (x + y)2= x2+ 2xy + y2

Thus, (x + y)2– 4 = x2+ 2xy + y2– 4

Therefore, the answer isx^2+ 2xy + y^2– 4

Question 16

is what you get when you multiply the same binomial to itself three times.

Answer 16

cube of a binomial

Question 17

The cube of a binomial (x + y) can be expressed as:

Answer 17

(x + y)^3= (x + y)(x + y)(x + y)

Question 18

To find the cube of a binomial:

Answer 18

Cube the first term of the binomial (or raise the first term to the exponent of 3).Multiply the square of the first term by the second term then multiply the product by 3.Multiply the first term by the square of the second term then multiply the product by 3.Cube the last term (or raise the last term to the exponent of 3).Combine the results you have obtained from Step 1 – 4.

You must also need to consider the operation used in the binomial. If it’s addition, all of the terms of the expansion are positive. On the other hand, if the operation is subtraction, the second and the last terms are negative and the rest are positive.

Question 19

Example 1:Expand (a + 1)^3

Answer 19

Solution:

Step 1: Cube the first term of the binomial (or raise the first term to the exponent of 3). The first term of the binomial isa. Cube ofais justa3.

Step 2: Multiply the square of the first term by the second term then multiply the product by 3.The first term of the binomial isaand its square is a2. We multiply a2to the second term which is 1. Hence, a2✕ 1 = a2. Finally, we multiply a2by 3 to obtain 3a2.

Step 3: Multiply the first term by the square of the second term then multiply the product by 3. The first term of the binomial which isamust be multiplied by the square of the second term which is 1 (12= 1). Thus, we havea✕1 = a. We then multiplyaby 3 to obtain 3a.

Step 4: Cube the last term (or raise the last term to the exponent of 3). The last term of the binomial is 1 and its cube is just 13= 1.

Step 5: Combine the results you have obtained from Steps 1 – 4. Combining what we have obtained from Steps 1 – 4, we have a3+ 3a2+ 3a + 1.

Since (a + 1)3has the addition sign, it means that the terms of the expansions must be all positive.

Therefore, (a + 1)^3=a^3+ 3a^2+ 3a + 1.

Question 20

Example 2:Expand (2a – b)^3

Answer 20

Solution:

Let us apply the steps on how to cube a binomial:

Step 1: Cube the first term of the binomial (or raise the first term to the exponent of 3). The first term is 2a and its cube is (2a)3= 8a3.

Step 2: Multiply the square of the first term by the second term then multiply the product by 3.The first term is 2a and its square is 4a2. We multiply the latter by the second term, which isb. Hence, 4a2✕ b = 4a2b . Then, we multiply the product by 3: 4a2b ✕ 3 = 12a2b.

Step 3:Multiply the first term by the square of the second term then multiply the product by 3. The first term of the binomial is 2a. We multiply 2a to the square of the second term (the second term isband its square isb2). Thus, 2a ✕ b2= 2ab2. Then, we multiply the product by 3 to give the following result: 2ab2✕ 3 = 6ab2

Step 4: Cube the last term (or raise the last term to the exponent of 3). The last term of the binomial isb, and the cube ofbisb3.

Step 5: Combine the results you have obtained from Steps 1 – 4

Since the binomial (2a – b) involves a subtraction sign, it means that the second and the last terms of the expansion must be negative and the remaining terms must be positive.

Therefore, the answer is8a^3– 12a^2b + 6ab^2– b^3.

Question 21

the process of determining the factors of a certain expression orpolynomial.

Answer 21

Factoring

Question 22

For example, the trinomial x2+ 4x + 3 can be factored as

Answer 22

(x + 1)(x + 3). Using the FOIL method, you can verify that (x + 1)(x + 3) = x^2+ 4x + 3

Question 23

Suppose we have 3x2and 6x.How can we determine the GCF of these expressions?

Answer 23

We need to performprime factorizationon these expressions. To do this, we write the numerical coefficients as a product of its prime factors and just write the variables with exponents in expanded form.

3x2= 3 ⋅ x ⋅ x

6x = 2 ⋅ 3 ⋅ x

To find the GCF of these expressions, we just take the common factors of the expressions and multiply them together. In our list above, note that the common factors are 3 andx. Thus, the GCF of 3x2and 6x is 3x.

Question 24

A quicker way to determine the GCF of monomials is by following these steps:

Answer 24

Find the GCF of the numerical coefficients.Obtain the common variables and write the smallest exponent among these common variables.Multiply what you have obtained from Steps 1 and 2. The result is the GCF of the monomials.

Question 25

Example 1:Let us try the above steps to find the GCF of 35y^2and 49y^3.

Answer 25

Solution:

Step 1: Find the GCF of the numerical coefficients.The GCF of 35 and 49 is 7.

Step 2: Obtain the common variables and write the smallest exponent among these common variables.The common variable between 35y2and 49y3isy. We put the smallest exponent among the common variables toy. Notice that the smallest exponent is 2. So we put 2 as the exponent ofy. Thus, we have y2.

Step 3: Multiply what you have obtained from Steps 1 and 2. We have obtained 7 from Step 1 and y2from Step 2. Thus, the GCF is7y^2.

Question 26

Example 2:What is the GCF of -16x^2y^3and 2xy^4z?

Answer 26

Solution:

Step 1: Find the GCF of the numerical coefficients.The GCF of -16 and 2 is -2.

Step 2: Obtain the common variables and write the smallest exponent among these common variables.The common variables between the given monomials arexandy. The smallest exponent inxis 1 while the smallest exponent inyis 3. Thus, we have xy3.

Step 3: Multiply what you have obtained from Steps 1 and 2.We have obtained -2 from Step 1 and xy3from Step 2. Thus, the GCF is-2xy^3.

Question 27

Example 3:What is the GCF of a^4b^3c and ab^5c^2?

Answer 27

Solution:

Step 1: Find the GCF of the numerical coefficients. Both expressions have a numerical coefficient of 1. Thus, the GCF is 1.

Step 2: Obtain the common variables and write the smallest exponent among these common variables.The common variables area, b,andc. The smallest exponent ofais 1, the smallest exponent ofbis 3, and the smallest exponent ofcis 1. Thus, we have ab3c.

Step 3: Multiply what you have obtained from Steps 1 and 2.We have obtained 1 from Step 1 and ab3c from Step 2. Thus, the GCF is ab^3c.

Question 28

Let us try to factor 15x^2+ 3 using its GCF. Again, recall that factoring means determining the factors of a certain expression.

Answer 28

The first step is to determine the GCF of the terms of the polynomial. The terms of 15x2+ 3 are 15x2and 3. Thus, we need to find the GCF of 15x2and 3.

Using the steps we have learned earlier to find the GCF of monomials (since both 15x2and 3 are monomials), we will be able to obtain the GCF which is 3.

The next step is to divide each term of the polynomial by the GCF.

As you can see above, we’re able to obtain 5x2and 1. Combining them will give us 5x2+ 1. This means that the GCF (which is 3) and 5x2+ 1 are the factors of 15x2+ 3.

Therefore, if we factor 15x2+ 3, we have3(5x2+ 1). You can verify using thedistributive propertythat 3(5x^2+ 1) = 15x2+ 3.

Question 29

Example 1:Factor 14a^2b^3– 32a^3b

Answer 29

Solution:

The GCF of the terms of the given polynomial is2a2b.

We divide the terms of the given polynomial by the GCF:

By dividing the terms by the GCF, we’re able to obtain 7b2and -16a. Combining them will give us 7b2– 16a.

This means that the GCF (which is 2a2b) and 7b2– 16a are the factors of 14a2b3– 32a3b

Therefore, if we factor 14a2b3– 32a3b, we’ll have2a2b(7b2– 16a). You can verify using the distributive property that 2a^2b(7b^2– 16a) = 14a2b3– 32a3b

Question 30

Example 2:Factor -21pq^2r + 9pqr

Answer 30

Solution:

The GCF of the terms of the given polynomial is -3pqr.

Dividing each term of the given polynomial by -3pqr:

We have obtained 7q and -3. Combining them will give us 7q – 3.

Thus,-21pq2r + 9pqr = –3pqr(7q – 3).

Question 31

Example 3:Factor ab + ad

Answer 31

Solution:

The GCF of the terms of the given polynomial isa.

If we divide each term of the polynomial bya, we will obtainbanddrespectively. Combining them will give usb + d.

Therefore, ab + ad =a(b + d).

Question 32

Example 4:Factor -12x^2+ 9x + 3 using the Greatest Common Factor.

Answer 32

Solution:

The GCF of the terms of the given polynomial is -3.

Dividing each term of the polynomial by -3:

We have obtained 4x2, -3x, and -1. Combining them will give us 4x2-3x – 1.

Therefore, -12x2+ 9x + 3 =-3(4x^2-3x – 1).

Question 33

a trinomial in the form ax2+ bx + c, wherea, b,andcare real numbers butais not equal to 0

Answer 33

quadratic trinomial

Question 34

Example 1:Is 8x^2+ 2x + x a quadratic trinomial?

Answer 34

Solution: No. Note that we can combine2xandxsince they are like terms. Thus, 8x2+ 2x + x is actually 8x2+ 3x. 8x2+ 3x, although it is quadratic, is not a trinomial.

Question 35

Let us begin with the simplest type of quadratic trinomials: those whose leading coefficient is 1 such as x2+ 5x + 6.

Answer 35

A quadratic trinomial can be obtained when we multiply two binomials by the FOIL method. Hence, if we factor a quadratic trinomial, there should be two binomials.

Here are the steps to factor a quadratic trinomial ifa = 1:

Write the binomials with the first terms as the square root of the leading term of the given quadratic trinomial.Think of the factors of the third term whose sum is equal to the second term.Write the numbers you have obtained from Step 2 as the second term of the binomials.

Question 36

Example 1:Let us apply these steps to factor x^2+ 5x + 6.

Answer 36

Solution:

Step 1: Write the binomials with the first terms as the square root of the leading term of the given quadratic trinomial.We start by writing two binomials withxas the first terms.

Step 2: Think of the factors of the third term whose sum is equal to the second term. The third term of x2+ 5x + 6is 6. Think of the factors of 6 that will give you the second term (which is 5).

Here are the factors of 6 and their sums:

1 and 6 (sum: 1 + 6 = 7)-6 and -1 (sum: -1 + (-6) = -7)3 and 2 (sum: 3 + 2 = 5)-3 and -2 (sum: (-3) + (-2) = -5)

Looking at pairs of factors of 6 above, the sum of 3 and 2 is equal to the second term which is 5.

Step 3: Write the numbers you have obtained from Step 2 as the second term of the binomials.In this case, 3 and 2 are the factors of 6 we obtained from Step 2 so these numbers become the second term of the binomials:

Thus, the factored form of x2+ 5x + 6 is(x + 3)(x + 2).

Question 37

Example 2:Factor x^2+ 8x + 15.

Answer 37

Solution:

Step 1: Write the binomials with the first terms as the square root of the leading term of the given quadratic trinomial.We start by writing two binomials withxas the first terms.

Step 2: Think of the factors of the third term whose sum is equal to the second term.The third term of x2+ 8x + 15is 15. Think of the factors of 15 that will give you the second term (which is 8).

Here are the factors of 15 and their sums:

1 and 15 (sum: 1 + 15 = 16)-1 and -15 (sum: -1 + (-15) = -16)3 and 5 (sum: 3 + 5 = 8)-3 and -5 (sum: (-3) + (-5) = -8)

Looking at the pairs of factors of 15 above, the sum of 3 and 5 is equal to the second term which is 8.

Step 3: Write the numbers you have obtained from Step 2 as the second term of the binomials.In this case, the numbers 3 and 5 each becomes the second term of the binomials:

Thus, the factored form of x2+ 8x + 15 is(x + 3)(x + 5).

Question 38

Example 3:Factor x^2– 12x + 27.

Answer 38

Solution:

Step 1: Write the binomials with the first terms as the square root of the leading term of the given quadratic trinomial.We start by writing two binomials withxas the first terms:

Step 2: Think of the factors of the third term whose sum is equal to the second term.The third term of x2– 12x + 27 is 27. Think of the factors of 27 that will give you the second term (which is -12).

Here are the factors of 27 and their sums:

9 and 3 (sum: 9 + 3 = 12)-9 and -3 (sum: -9 + (-3) = -12)27 and 1 (sum: 27 + 1 = 28)-27 and -1 (sum: (-27) + (-1) = -28)

Looking at the pairs of factors of 27 above, the sum of -9 and -3 is equal to the second term which is -12.

Step 3: Write the numbers you have obtained from Step 2 as the second term of the binomials.In this case, the numbers -9 and -3 we have obtained from Step 2 become the second term of the binomials:

Thus, the factored form of x2– 12x + 27 is (x + (-9))(x + (-3)) or more appropriately,(x – 9)(x – 3).

Question 39

Here are the steps to factor a quadratic trinomial ifa≠ 1

Answer 39

Multiply the coefficients of the first and third terms of the quadratic trinomial.Think of the factors of the number you have obtained in Step 1 whose sum is equal to the coefficient of the second term.Expand the second term of the trinomial using the factors you have obtained from Step 2. After expanding, the expression should now consist of four terms.Group the trinomial into two groups.Factor out the GCF of each group. Once you have factored out the GCF, expect that there will a common binomial.Factor out the the common binomial

Question 40

Example 1:Factor 3x^2– 4x + 1

Answer 40

Solution:

Step 1: Multiply the coefficients of the first and third terms of the quadratic trinomial.The coefficient of the first term is 3 while the coefficient of the third term is 1. Multiplying these two numbers to each other will give us the number 3 as the product.

Step 2: Think of the factors of the number you have obtained in Step 1 whose sum is equal to the coefficient of the second term.The number we have obtained from Step 1 is 3. Think of the factors of 3 such that their sum is the coefficient of the second term which is -4.

Here are the factors of 3 together with their sums:

3 and 1 (sum is 4)-3 and -1 (sum is -4)

As you can see, the factors of 3 that give -4 as their sum are -3 and -1.

Step 3: Expand the second term of the trinomial using the factors you have obtained from Step 2. After expanding, the expression should now consist of four terms.The second term of the trinomial 3x2– 4x + 1 is -4x. We will expand it by replacing it with the numbers we have obtained from Step 2. Recall that we have obtained -3 and -1 from Step 2. Thus, we will replace -4x with -3x and -1x:

3x2– 3x – 1x + 1

Step 4: Group the trinomial into two groups.We group the trinomials using parentheses:

(3x2– 3x) – (1x + 1)

Step 5: Factor out the GCF of each group. Once you have factored out the GCF, expect that there will be a common binomial.

We have(3x2– 3x) – (1x + 1). The first group is 3x2– 3xwhile the second group is 1x + 1 or x + 1. The GCF of the first group is 3x while the GCF of the second group is 1. We factor out the GCF of the respective groups:

3x(x – 1) – 1(x – 1)

Note that (x – 1) is the common binomial to 3x(x – 1) – 1(x – 1).

Step 6: Factor out the common binomial.In 3x(x – 1) – 1(x – 1), (x – 1) is the common binomial. We factor it out to complete the factoring process.

(x – 1)(3x – 1)

Therefore, the factored form of 3x2– 4x + 1 =(x – 1)(3x – 1)

Here’s a preview of what we have performed above:

3x2– 4x + 1

3x2– 3x – x + 1

(3x2– 3x) – (x + 1)

3x(x – 1) – 1(x – 1)

(3x – 1)(x – 1)

Question 41

Example 2:Factor 2x^2+ 5x + 2.

Answer 41

Solution:

Step 1:Multiply the coefficients of the first and third terms of the quadratic trinomial.The coefficient of the first term is 2 while the coefficient of the third term is 2 as well. Their product is 4.

Step 2:Think of the factors of the number you have obtained in Step 1 whose sum is equal to the coefficient of the second term. The number we have obtained from Step 1 is 4. Think of the factors of 4 such that their sum is the coefficient of the second term which is 5.

Here are the factors of 4 together with their sums:

4 and 1 (sum is 5)-4 and -1 (sum is -5)2 and 2 (sum is 4)-2 and -2 (sum is -4)

As you can see, the factors of 4 that give 5 as their sum are 4 and 1.

Step 3: Expand the second term of the trinomial using the factors you have obtained from Step 2. After expanding, the expression should now consist of four terms.The second term of the trinomial 2x2+ 5x + 2 is 5x. We will expand it by replacing it with the numbers we have obtained from Step 2. Recall that we have obtained 4 and 1 from Step 2. Thus, we will replace 5 with 4 and 1:

2x2+ 4x + x + 2

Step 4: Group the trinomial into two groups.We group the trinomials using parentheses:

(2x2+ 4x) + (x + 2)

Step 5: Factor out the GCF of each group.Once you have factored out the GCF, expect that there will be a common binomial.

Continuing from the previous step, we have (2x2+ 4x) + (x + 2). The first group is 2x2+ 4x while the second group is x + 2. The GCF of the first group is2xwhile the GCF of the second group is 1. We factor out the GCF of the respective groups:

2x(x + 2) + 1(x + 2)

Note that (x + 2) is the common binomial to 2x(x + 2) + 1(x + 2).

Step 6: Factor out the the common binomial.In 2x(x + 2) + (x + 2), (x + 2)is the common binomial. We factor it out to complete the factoring process.

(x + 2)(2x + 1)

Therefore, the factored form of 2x2+ 5x + 2 =
(x + 2)(2x + 1)

Question 42

a quadratic trinomial that is derived from squaring a binomial.

Answer 42

Perfect square trinomial

Question 43

Example:Is x^2+ 6x + 9 a perfect square trinomial?

Answer 43

Yes

Question 44

Once you have confirmed that a quadratic trinomial is a perfect square trinomial, you can factor it using the steps below:

Answer 44

Get the square root of the first term. It is the first term of our factors.Get the positive square root of the last term. It is the second term of our factors.If the perfect square trinomial has a subtraction sign, then the factors will have a subtraction sign. Otherwise, they will be using an addition sign.

Question 45

Example 1:Factor x^2+ 14x + 49.

Answer 45

Solution:

Step 1: Get the square root of the first term. It is the first term of our factors.The first term isx2and its square root isx. Thus,xis the first term of our factors.

Step 2: Get the positive square root of the last term. It is the second term of our factors.The last term is 49 and its square root is 7. Thus, 7 is the last term of our factors.

Important Note: When we get the square root of a number, we actually obtain two values–one is a positive and the other is a negative number. For instance, the square root of 49 is 7 and -7. However, in this case of factoring perfect square trinomials, we will only consider the positive square root of the third term.

Step 3:If the perfect square trinomial has a subtraction sign, then the factors will have a subtraction sign. Otherwise, they will be using an addition sign.

Sincex2+ 14x + 49 has no subtraction sign involved, then the binomials do not have a subtraction sign but an addition sign instead.

Thus, x2+ 14x + 49 =(x + 7)(x + 7)

Question 46

Example 2:Factor x^2– 18x + 81.

Answer 46

Solution:

Step 1:Getthe square root of the first term. It is the first term of our factors.

Square root ofx2isx.

Step 2:Get the positive square root of the last term. It is the second term of our factors.

Square root of 81 is 9.

Step 3:If the perfect square trinomial has a subtraction sign, then the factors will have a subtraction sign. Otherwise, they will be using an addition sign.

Thus, x2– 18x + 81 =(x – 9)(x – 9)

Question 47

binomial in the form a2– b2

Answer 47

Difference of two square

Question 48

To factor a difference of two squares:

Answer 48

Get the square root of the first term and the square root of the last term.Express the factors as the sum and difference of the quantities you have obtained in Step 1.

Question 49

Example:Factor a^2– 9.

Answer 49

Solution:

Step 1: Get the square root of the first term and the square root of the last term.The square root of the first term isawhile the square root of the second term is 3.

Step 2: Express the factors as the sum and difference of the quantities you have obtained in Step 1.Expressing the quantities we have obtained from Step 1 as sum and difference, we have (a – 3)(a + 3).

Thus, the answer is(a – 3)(a + 3).

Question 50

1) What is (x + 3)(x - 1)?
a) x^2 + 2x + 3
b) x^2- 2x - 3
c) x^2 + 2x - 3
d) x^2- 2x + 3

Answer 50

C

Question 51

2) Expand (2x - z)^2

a) 4x^2- 4xz + z^2
b) 4x^2z - 4xz + z^2
c) 4x^2- 2xz + z^2
d) 4x^2- 4xz + 2z^2

Answer 51

A

Question 52

3) What is (a - p)(a + p)
a) a2 + 2p2
b) a2- p2
c) 2a2- 2p2
d) a - p

Answer 52

B

Question 53

4) Factor x2- x - 12

a) (x - 4)(x - 3)
b) (x + 4)(x - 3)
c) (x + 4)(x + 3)
d) (x - 4)(x + 3)

Answer 53

D

Question 54

5) Factor 2a2b - 6ab

a) 2ab(a - 3)
b) 2ab(a - 3b)
c) 2ab(a2- 3)
d) 2a(ab - 3)

Answer 54

A