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| | 10 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Consider the following algebraic expressions. (n+2)^2-4 and n^2+4n These algebraic expressions are equivalent.
There can be different methods for factoring a quadratic expression, depending on its type. If the expression is the square of a binomial, it can be factored as a perfect square trinomial.
A perfect square trinomial is a trinomial that can be written as the square of a binomial. There are two general types of perfect square trinomials that can be useful when dealing with quadratic functions or quadratic equations.
| Perfect Square Trinomial | Square of Binomial |
|---|---|
| a^2+2 a b+ b^2 | ( a+ b)^2 |
| a^2-2 a b+ b^2 | ( a- b)^2 |
A perfect square trinomial should have one of the forms shown in the table — that is, the first and the last terms are perfect squares and the middle term is two times the product of the square roots of the first and last terms.
For a trinomial to be factorable as a perfect square trinomial, the first and last terms must be perfect squares and the middle term must be two times the square roots of the first and last terms. Consider the following expression. 16x^2+88x+121 To factor this trinomial, there are three steps.
Jordan has two copies of the same square painting, and she wants to frame these copies. The area of each painting is represented by a quadratic expression.
Before preparing the frames, Jordan needs to find the side length of each painting. What are the side lengths of the paintings in terms of x?
| Expression | First Term | Last Term |
|---|---|---|
| 49x^2-56x+16 | 49x^2=( 7x)^2 | 16=( 4)^2 |
| 16x^2+24x+9 | 16x^2=( 4x)^2 | 9=( 3)^2 |
Since the first and last terms are perfect squares, the trinomials are probably perfect square trinomials. To be sure, check whether the middle terms are two times the square roots of the first and last terms.
| Expression | First Term | Last Term | Middle Term |
|---|---|---|---|
| 49x^2-56x+16 | 49x^2=( 7x)^2 | 16=( 4)^2 | 56x=2* 7x* 4 |
| 16x^2+24x+9 | 16x^2=( 4x)^2 | 9=( 3)^2 | 24x=2* 4x* 3 |
The expressions satisfy the conditions to be a perfect square trinomials. From here, each trinomial can be written as a square of a binomial. 49x^2-56x+16 & = ( 7x- 4)^2 16x^2+24x+9 & = ( 4x + 3)^2 By taking the square roots of these expressions, the side lengths of the paintings can be found. Side Length of Larger Painting sqrt(( 7x- 4)^2) = 7x- 4 [1em] Side Length of Smaller Painting sqrt(( 4x + 3)^2) = 4x + 3
Ali and Jordan are factoring the following quadratic expression. 121m^2l ^2-44ml c+4c^2 However, they obtained different results.
Therefore, it would be a good idea to start by checking if the given expression is indeed a perfect square trinomial. To do this, check if the first and last terms of the expression are perfect squares. 121m^2l ^2-44ml c+4c^2 ⇓ ( 11ml )^2-44ml c+( 2c)^2 As it can be seen, the first and last terms are perfect squares. This means that the given expression could be a perfect square trinomial. To be sure, check whether the middle term is two times the square roots of the first and last terms. ( 11ml )^2-44ml c+( 2c)^2 ⇓ ( 11ml )^2-2( 11ml )( 2c)+( 2c)^2 Since the expression satisfies the conditions, it is a perfect square trinomial and can be written as a square of a binomial. 121m^2l ^2-44ml c+4c^2 ⇓ ( 11ml )^2-2( 11ml )( 2c)+( 2c)^2 ⇓ ( 11ml - 2c)^2 With this information, it can be concluded that Jordan is correct. On the other hand, Ali only took the square root of the coefficients of the first and last terms, so he did not get the correct answer.
Note that the product of two conjugate binomials results in a difference of two squares. Therefore, the difference of two squares can be factored out using reverse thinking.
The product of a conjugate pair of binomials results in a difference of two squares. Using this relationship, the difference of two squares can be factored as the product of the sum and difference of two quantities. a^2-b^2=(a+b)(a-b) As an example, the following expression will be factored. 9x^2-121 There are two steps to factor the expression as a difference of two squares.
The volume of a rectangular prism is given by the expression 81x^4-16.
Recall that the volume of a rectangular prism is the product of its dimensions. Find the possible dimensions of the prism if the dimensions are binomials with integer coefficients.
Split into factors
a* a=a^2
a^m* b^m=(a * b)^m
The given expression is also equal to the square of 9x^2 minus the square of 4. From here, the expression can be written as the product of a conjugate pair of binomials. V=( 9x^2)^2-( 4)^2 ⇓ V=( 9x^2+ 4)( 9x^2- 4) The volume of the prism can be thought of as the product of its base area and height. Notice that the binomial 9x^2+4 cannot be factored further. However, the other binomial is also a product of a conjugate pair of binomials and can therefore be further factored as a difference of two squares.
Split into factors
a* a=a^2
a^m* b^m=(a * b)^m
Therefore, (9x^2-4) can be written as the product of the binomials 3x+2 and 3x-2. 81x^4 -16 ↙ ↘ (9x^2+4)(9x^2-4) ↙ ↘ (9x^2+4)(3x+2)(3x-2) The dimensions of the rectangular prism are 9x^2+4, 3x+2, and 3x-2.
Tadeo's older brother is in college. He currently lives in a one-room square apartment, from which a small square is cut out as shown in the diagram.
He wants to move to a rectangular room with the same area as his current room.
Therefore, the area of the larger square and the area of the smaller square need to be found first. Recall that area of a square is the square of its side length.
| Square | Side Length, s | Area, s^2 |
|---|---|---|
| Larger Square | 4n+1 | ( 4n+1)^2 |
| Smaller Square | 5 | 5^2 |
By finding the difference between the areas, the area of the room can be found. ( 4n+1)^2 - 5^2 Note that this is an example expression. An equivalent expression can also be written by expanding the squares.
a^2-b^2=(a+b)(a-b)
Add and subtract terms
Therefore, 4n-4 and 4n+6 should be the dimensions of the rectangular room. Since 4n+6 is greater than 4n-4, 4n+6 is the length of the room. Length: & 4n+6 Width: & 4n-4
Two square windows of a house and their areas are shown in the given image.
Write an expression that represents the difference between the areas of the windows. Show two different ways to find the solution.
& 25x^2-30x+9 - & x^2-14x+49 & 24x^2-16x-40 The resulting expression represents the difference of the areas.
| Expression | First Term | Last Term |
|---|---|---|
| 25x^2-30x+9 | 25x^2=( 5x)^2 | 9=( 3)^2 |
| x^2-14x+49 | x^2=( x)^2 | 49=( 7)^2 |
Since the first and last terms are perfect squares, there is a good chance that these expressions are perfect square trinomials. Now, check whether the middle terms are two times the square roots of the first and last terms.
| Expression | First Term | Last Term | Middle Term |
|---|---|---|---|
| 25x^2-30x+9 | 25x^2=( 5x)^2 | 9=( 3)^2 | 30x=2* 5x * 3 |
| x^2-14x+49 | x^2=( x)^2 | 49=( 7)^2 | 14x=2* x * 7 |
The expressions satisfy the conditions to be a perfect square trinomial. From here, they can be written as a square of a binomial. 25x^2-30x+9 & = ( 5x- 3)^2 x^2-14x+49 & = ( x - 7)^2 Since the expressions are in factored form, the difference in the areas can be found. (5x-3)^2 - (x-7)^2 As can be seen, the resulting expression is a difference of two squares. This can be factored further and written as a product of a conjugate pair of binomials.
a^2-b^2=(a+b)(a-b)
Distribute -1
Add and subtract terms
The final form of the expression can be obtained by multiplying the binomials.
The result is the same as subtracting the original polynomials.
Throughout the lesson, two methods have been covered for factoring special products. These methods together can be used to solve the challenge presented at the beginning of the lesson. Recall that there were two algebraic expressions. n^2+4n and (n+2)^2-4 These expressions are known to be equivalent.
The number of dots in each figure in terms of n can be obtained in two different ways, one represented by n^2+4n and the other represented by (n+2)^2−4. To illustrate n^2+4n, assume that n^2 is the inside full square of dots and 4n is the four outside borders with n dots each.
Notice that the number of dots in the outside border and side length of the inside square, in terms of n, match the figure number. Now, to illustrate (n+2)^2-4, imagine the larger square with the four additional dots filled in at the corners. Then, (n+2)^2 would be the number of dots in the larger square, since the missing 4 dots were added.
As it can be seen, the number of dots in n^(th) figure can be represented by n^2+4n and by (n+2)^2-4. Therefore, these expressions are equivalent.
Write as a power
a^2-b^2=(a+b)(a-b)
Add and subtract terms
Distribute n
Multiply
As shown, the expressions are equivalent.
To factor the given expression, we will recall the formula for the difference of two squares. a^2- b^2=( a+ b)( a- b) Now, we will rewrite our terms as a second power. Then we will apply the formula.
Notice that the given expression can be written as the product of two conjugate binomials, (x+9) and (x-9).
This time we will rewrite 25y^2-1 in the form a^2- b^2. Let's first write 25y^2 as ( 5x)^2 and then we need also to rewrite 1 as 1^2.
The given expression is equal to the product of conjugate binomials (5y+1) and (5y-1).
Finally, we will factor the expression 7-0.25z^2. We can do this by finding the number that is equal to 7 when it is multiplied by itself.
sqrt(7) * sqrt(7) = 7
Having the fact that the second power of sqrt(7) is equal to 7, we can rewrite our expression.
The expression is equal to the product of conjugate binomials (sqrt(7)+0.5z) and (sqrt(7)-0.5z).
Let's recall that the formula for the difference of two squares is a^2- b^2=( a+ b)( a- b). The given equation is written in that same form. 16y^2-△=( ◊ + 7)( 4y- ♡) With this in mind, if we match the first terms of the binomials, we can find that ◊= 4y. Then, if we match the second terms of the binomials, we can find that 7= ♡. 16y^2-△=( 4y + 7)( 4y- 7) We are almost done! Since the squared terms a^2 and b^2 in the left hand side are the squares of the terms inside the binomials, we can write the the values of a and b correspondingly. a^2- b^2=( a+ b)( a- b) ⇓ ( 4y)^2- 7^2=( 4y+ 7)( 4y- 7) Great! Now, we can complete our equation to have it in the given form. 16y^2-△=( ◊ + 7)( 4y- ♡) ⇓ 16y^2-49=( 4y+ 7)( 4y- 7) Let's finally match our all missing values. △=49, ◊=4y, and ♡=7 Therefore the answer is option B.
This time we will start by recalling the formula to factor a perfect square trinomial.
a^2+2 a b+ b^2=( a + b)^2
With this in mind, let's now rewrite and compare our equation.
a^2+2 a b+ b^2=( a + b)^2
⇓
( 3x)^2+△+ 5^2=( ◊ + ♡)^2
As we can see, we can find the values of a and b such that a= 3x and b= 5. Since we can match a with ◊ and also b with ♡, we can substitute that ◊= 3x and ♡= 5.
a^2+2 a b+ b^2=( a + b)^2
⇓
( 3x)^2+â–³+ 5^2=( 3x + 5)^2
Now, we will find the value of â–³ by substituting a= 3x and b= 5 into the middle term 2 a b.
Finally, we can write our all missing values. △=30x, ◊=3x, and ♡=5 Therefore, the answer is option C.
Factor the following trinomial. c^2+10c+25
We want to factor the given trinomial. c^2+10c+25 How do we know that the expression is a perfect square trinomial? Well, let's ask a few questions.
| Is the first term a perfect square? | c^2= c^2 ✓ |
| Is the last term a perfect square? | 25= 5^2 ✓ |
| Is the middle term twice the product of 5 and c? | 10c=2* 5* c ✓ |
As we can see, the answer to all three questions is yes! Therefore, we can write the trinomial as the square of a binomial. Note there is an addition sign in the middle.
Factor the following expression completely. 3k^4-3
To factor the given expression, we will first identify and factor out the greatest common factor. We will then use the formula for the difference of two squares to finish the problem.
The greatest common factor (GCF) of an expression is a common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest exponent. The GCF of the given expression is 3.
In this part of the solution, we will have a look closely at the expression k^4-1. It can be expressed as the difference of two perfect squares.
Recall the formula to factor a difference of squares. a^2- b^2 ⇔ ( a+ b)( a- b) We can apply this formula to our expression. 3 ( ( k^2)^2- 1^2 ) ⇔ 3( k^2+ 1)( k^2- 1) Next, let's take a look at the expression k^2-1. Once again, we can express it as a difference of two perfect squares. Let's do it!
Now, we can apply the formula to factor our expression completely. 3 (k^2+1)( k^2- 1^2) ⇕ 3(k^2+1)( k+ 1)( k- 1)
Solve the following equation. h^2-8/9h+16/81=0
We will solve the given equation by factoring. To do so, we will start by identifying its first term and last term to check whether it is a perfect square trinomial or not. h^2-8/9h+16/81=0 ⇕ h^2-( 8/9 )h+ 16/81=0 Notice that the first term h^2 and the last term 1681 are the perfect squares. Let's see whether the middle term can be written as 2 a b to be able to say it is a perfect square trinomial. h^2-( 8/9 )h+ 16/81=0 ⇕ h^2-2 * h * ( 4/9 )+( 4/9 )^2=0 Great! As it is seen, the given trinomial can be rewritten in the form a^2-2 a b+ b^2. Therefore, it is a perfect square trinomial. Let's now factor it by applying the formula.
Now we are ready to use the Zero Product Property.
We found that the solution to the given equation is h= 49.