6.4 Add/Subtract Rational Expressions with Binomial and Trinomial Denominators

Adding and subtracting RE with binomial and trinomial denominators involves using the skills of factoring (when needed), identifying the LCD, and the NPV. Adding and subtracting RE with binomial and trinomial denominators uses the same characteristics as a monomial denominator, the goal is to have the same/common denominators in order to add or subtract the numerators. Here is a simple expression to simplify,

As you can see the denominators in this expression are the same which means we can just add the numerators which is 4x + 6x and the answer is

We can also determine the NPV (non-permissible value), which is -5 (x≠-5). NPV is important because it makes us become aware of which values are not allowed into the denominator, because if the x in the denominator equals -5 then the denominator would equal to 0 (-5+5=0) and we know that a number cannot be divided by zero. If you put any number divided by 0 in your calculator it would should up as ERROR.

Here are the steps for adding and subtracting rational expressions:

1. Factor the expression
2. Identify the LCD (Lowest Common Denominator)
3. State the NPV (Non-Permissible Value)
4. Rewrite fraction over the LCD

Here is another simple example but with different denominators:

We cannot factor but we can identify the LCD which is:

Since the denominator on the left fraction is missing an (x) we must times (x) by the numerator and denominator and with the right fraction it is missing a (z) so you times the numerator and denominator by z so the denominators would be the same. It is important to remember that what happens to the denominator must happen with the numerator as well.

Then, state the NPV which are x ≠ 0 and z ≠ 0

Since the denominators are the same we can combine them and make it one fraction (over the LCD)

Here is another equation:

As you can see the denominators are different so they must be the same in order to simplify, you can do the cross method where the left denominator multiplies with the right numerator and the right denominator multiplies with the left numerator, after multiplying the numerators remember to multiply the denominators as well, it should look something like this:

Note how the denominators are the same, now we can add m to 5m and -4 to 15 which makes our answer to:

and the NPV is m ≠ -3, 4

Here is an equation that requires factoring to start:

First, we factor the denominators in both fractions

We can then switch the right fraction denominator into -x(x-6), this would make finding and calculating the LCD easier and simpler

Then identify the LCD which is x(x-6)(x+6) and the NPV are x ≠ 6,-6,0

Remember that subtracting a negative fraction is the same as adding a positive fraction so we can rewrite the equation as

Then we apply the LCD to the fractions

Then the numerator becomes 2x+6, which we can factor to 2(x+3)

We want to factor the numerator because in some cases, you can actually simplify a part of the expression in the numerator with the denominator which can help you fully simplify your expression.

Here is one more binomial expression that involves expanding trinomials:

Since we cannot factor we can identify the LCD which is (t-4)(t-5). So in order for the LCD to become (t-4)(t-5), is on the left fraction you must multiply both numerator and denominator by (t-5), it is (t-5) because the numerator already has the    (t-4) and the LCD is (t-4)(t-5) same with the other fraction but instead multiplying by (t-5) it is (t-4). The NPV is t ≠ 4,5

Then the numerator would be t²-2t-15 – (t²-6t+8), it is very important that you keep the trinomial in brackets because of the negative symbol, because if it is just t²-2t-15 – t²-6t+8, then that is wrong because the negative only applied to t². If there are brackets then the negative would apply to every number so t²-2t-15 – (t²-6t+8) would turn to t²-2t-15 – t²+6t-8,

next we can collect like terms which makes it 4t-23 so the answer would be:

Here is an example that include a trinomial in its denominator:

We can first factor x²-2x-8, (you will need to acquire the skill of factoring trinomials in order to correctly simplifying adding/subtracting rational expressions)

Here is the factor form of x²-2x-8, also I personally like to put the 3 over 1 so I can visualize multiplying the denominator.

Then we find the LCD which is (x-4)(x+2), which the right fraction already has so you just multiply (x-4)(x+2) to the left fraction.

The NPV would be x ≠ -2,4

Then you can expand the left fraction numerator

Then we can write the expression over the LCD and combine like terms

Then factor numerator

Here is one last example that seems intimidating but if you use the same skills as before it is quite simple

First factor all the trinomials in the denominator

Then we can change the middle fraction to addition by multiplying the denominator by negative 1

Then apply LCD to the numerators according to each numerator. The LCD is (x+3)(x-2)(x+2). So the NPV is x≠ -3.-2, 2

Expand numerators

Collect like-terms and write fraction over LCD

Factor numerator

As you can see there is (x+2) and (x+3) in the numerator and denominator so they can cancel out

Which makes our final answer to:

NPV = x ≠ -3.-2, 2

It is important to find your non-permissible values after you find your LCD because as you can see the values can disappear when fully factored

At the end if you want to verify if your fully simplified version matches your original expression just pick any number(s) that isn’t a NPV and plug it into the variables in the expression and if you get the same answer in both simplified and non simplified version then you simplified it correctly

In conclusion, adding and subtracting rational expression with binomial and trinomial denominators is quite easy as long as you have the foundation skills of doing them, such as, factoring trinomials/binomials, identifying the LCD, stating the non-permissible value, and a general sense of combing like-terms regarding the numerator and denominator.

3.3 Solving Quadratics by Factoring

First, if we want to begin solving quadratics by factoring, we must know how to factor as it is the basics of solving quadratic equations. For example, if you want to solve (2x² – 15x + 25 = 0), we must begin with factoring this equation, and that is by identifying the product of A and C and finding what two numbers make that product of A and C that also is the sum of B. Such as, in this case, A = 2, B = -15, and C =25, there are two methods in factoring this which is either the X method or the square box method, personally I prefer the X method. The factors of 50 (because A • C = 2 • 25 = 50) is 1 • 50, 2 • 25, and 5 • 10. Which one of these has a difference or sum of -15? 5 and 10, in this case, it would be -5 and -10 because (-5 • -10 = our (ac) value of 50 as well as it is the sum of our (b) value of -15 (-5 – -10 = -15).) Next, once we have these values we can use the X method to factor,

after we have placed our values, we can divide the left and right side by the A value

Yes the 1 underneath the 5 is unnecessary but personally, it helps me visualize the factor more easily. Then, to factor it is (2x-5)(x-5), personally, how I usually distinguish which numbers go where is if you look at the numerator of each fraction which is (-5 and -5) you can recognize the plus or minus symbol which is a minus in this case and if you compare it to the recipe of a factored quadratic which looks like (x -/+ #)(x -/+ #), you can sort of visualize the placement of the numerator value into a factored form from matching the symbol (+ -) placement. Another way to remember is bottoms up which is to find the denominator (bottom) and moving it up which makes it first and before the numerator when writing them into two factors in brackets. After that, we can find which values would make it so the equation would equal 0. For (2x-5)(x-5) X = 5/2 and X = 5 because (2 (5/2) – 5 = 0) and (5-5 = 0)

Once we know how to factor we can finally begin solving quadratics. A question that looks different but uses the same formula of factoring quadratics is (2x – 3)(x + 1) = 3. First, the right side must equal 0 and the left side must be in general form (Ax²+Bx+C), so expand the left side and simplify,

2x² + 2x – 3x -3 -3 = 0

2x² – x – 6 = 0

Factor trinomial

(2x + 3)(x – 2) = 0

Then, to solve, 2x + 3 = 0 -> 2x = -3 -> x = -3/2

AND x – 2 = 0 -> x = 2

After we can verify the solution,

(2(2) -3)(2+ 1) = 3

(1)(3) = 3

3 = 3

AND

(2(-3/2) – 3) (-3/2 +1) = 3

(-6)(-1/2) = 3

3 = 3

Next, we can try an equation that involves a common factor, one example could be (2x² + 18 = 12x)

First we must move everything to the left side so the equation could equal 0

(2x² -12x +18 = 0)

Then remove GCF which is 2 because the value of A, B, C in the equation can be divided by 2 (2,-12,18)

2(x² -6x + 9) = 0

Then factor

2(x – 3) (x – 3) = 0

X = 3, X = 3

Verify

2(3)² + 18 = 12(3)

18 + 18 = 36

36 = 36

One more simple example that is not quite the general form but still uses the same properties of solving quadratic equations is

(2x²=4x)

Move 4x to the left side to make the equation equal to 0

(2x² – 4x = 0)

Factor the polynomial

(2x (x – 2) = 0)

X = 2 and X = 0 (x = 0 because if you put 0 in the x in 2x then it would be 2(0) and 2 •  0 = 0)

Verify

2(2)² = 4(2)

8 = 8

AND

2(0)² = 4(0)

0 = 0

You can also apply this strategy of solving quadratic with radical equations, such as (√23-x = x – 3), we must rewrite this equation into Ax²+Bx+C = 0 form, to do that we first isolate the radical, which makes the equation to,

(√23-x)² = (x – 3)²

we square both sides to remove radical

(23 – x = (x – 3)(x – 3) )

Then expand the right side

(23 – x = x² -6x + 9)

Move the left side to the right side to make the equation equal 0

(x² – 5x – 14 = 0)

Factor

(x – 7) (x + 2)

X = 7, and x = -2

Verify

(√23- (7) = (7) – 3)

(√16 = 4)

4 = 4

AND

(√23- (-2) = (-2) – 3)

(√25 = -5)

5 ≠ -5

As you can see all the questions before this the solution worked when verifying, however, when it comes to radicals the solution may not always work, which is why it is very important to verify especially when dealing with radicals. This means that (-2) in the example above is an extraneous root, where it may seem like the solution would work because you factored it, but it actually doesn’t and is false.

An example that uses the method of solving quadratic but also incorporates a method of the previous chapter of substituting variables is (2x-1)² – 2(2x-1) – 8 = 0, this looks like Ax² + Bx + C = 0, we can replace (2x-1) to W, which makes the equation w² – 2w – 8 = 0 then we factor, (w-4)(w+2), then we replace (w) with (2x-1), which equals to (2x-1-4)(2x-1+2), simplify, (2x-5)(2x+1), then x = 5/2 and x = -1/2, verify, also when verifying, make sure you input the solutions in the original equation which is (2x-1)² – 2(2x-1) – 8 = 0, to make sure the answers are accurate

X=5/2:(4)² – 2(4) – 8 = 0, 0 = 0, AND X=-1/2: (-2)² – 2(-2) – 8 = 0, 0 = 0.

Lastly, we can use a quadratic equation to solve a problem, this makes you think differently as you need to know which numbers are needed and where they are needed to be put, instead of just doing equations already given to you.

Such as, the area of a rectangular sheep pen is 96m². The pen is divided into two smaller pens by inserting a fence parallel to the width of the pen. A total of 48 m of fencing is used. Determine the dimensions of the pen.

The area of the pen is 96m² and the width is W in metres, to find the length it would be 96/W = L, which are the dimensions. The total length of the fence is 48m so an equation would be (3w + 2(96/w) = 48) First we multiply each term by w to remove the w underneath the 96, (3w² + 2(96) = 48w), make it general form, (3w² -48w + 192), next we divide each term by 3, 3(w² – 16 + 64), then factor, 3(w – 8)(w – 8), so (w = 8), lastly 96m²(area)/ 8m(width) = 12m(length), so the sheep pen has a length of 12m, a width of 8m, and a area of 96m²

In conclusion, solving quadratic by factoring is simple and some important things to remember and to do is when solving equations are, know how to factor quadratics, know how to make it into general form (Ax² + Bx + C= 0), verify the solution (especially radicals), and your main goal is to know X or what the variable equals to; to make the equation true.

Adding and Subtracting Radical Expressions:

To start we can remind ourselves how to simplify and work with like terms, the strategies for simplifying polynomials can be used to simplify sums and differences of radicals. For simplifying polynomials an example could be (3x + 5x =), since the variables are the same (x) then we can add the coefficients which equals to (8x). For radicals, what must be the same in order to simplify each other is the radicand and index, such as, (4√2 + 3√2) can be simplified to (7√2). Another example could be (5√2 + 3√18 – 4√8) Although it may seem like the radicands are different, you can simplify the radical (3√18) into (9√2) and (4√8) into (8√2) then we can simplify that into (5√2 + 9√2 – 8√2 = 6√2). Also, it is important to notice that the indexes are the same, which is 2 in this case, if not, you cannot simplify the radicals such as (3√3 + 4∛3) because even though they have the same radicand the indexes are different at 2 and 3. That being said, once we have like terms (radicand and index) we can simplify them by combining the co-efficient, and ultimately rewriting the equation as the most simplified result as it can be. To add, knowing the prime factors of numbers and how to simplify radicals are really important as it is basically the foundation of doing these types of questions like if you don’t know how to find the prime factors of a number you cannot fully simplify the radical making it really difficult to complete a question.

We can try a question that is a little harder but still builds with the knowledge and rules we know.

Simplify (√20x⁵y⁴ – √125xy²)  First we know the indexes are the same so that’s good but the radicands are not, so the next step is to simplify 20 and 125 which makes the expression into (2√5x⁵y⁴ – 5√5xy²), next are the variables, for (2√5x⁵y⁴) 2 groups of 2 of x⁵ can move to the co-efficient because the index is 2 which leaves 1 x remaining in the radicand, for y⁴ 2 groups of 2 can move with the co-efficient which is all the y’s because 2 x 2 = 4 and it is y⁴ (y⋅y⋅y⋅y). We can apply the same with (5√5xy²), in this case, only 1 group of 2 of y can move next to the co-efficient because it is y², the reason for being groups of 2 is due to the index (√ no number means 2) if it was ∛ then it would be groups of 3. After simplifying the radicand’s it brings it to (2x²y²√5x – 5y√5x) As you can see the radicands are the same so we can simplify to ( (2x²y² – 5y)√5x) ) TIP: If you are not sure if you did it right, just expand the simplified radicals and see if you got the original expression.

One harder question is (5√8x³ + 4y√75y³ – 2√27y⁵ – 3x√50x + 4∛3y)

First, simplify the numbers in the radicand

Then the variables

Combine like terms

Simplified!

Along with all this, we can also identify the values of the variable after simplifying and there are some rules, one rule is how the radicand cannot be negative if the index is even, for example, 5√x + 3√x – 4√x = 4√x, since the index is 2 the radicand cannot be negative, meaning each radical is defined for x ⩾ 0, or else it would be called “undefined” or “non-real number”. However, if the index is odd like 5∛x then it can be defined as x ∈ R (any real number). If it is something like q∜p³q, then to be defined, p⩾0 q⩾0 or p⩽0 or q⩽0, because if both p and q are negative in the radicand then the outcome would be positive which is why a negative could be possible in the radicand even with an even index as long as it comes out as a positive. Another scenario when there could be negatives in the radicand with an even index is if a question were to be 2√2a², this would be (a ∈ R) because of the exponent of 2, for example, if a = -2, it would become 2√2 (-2)(-2), which equals to positive 4 (2√2(4)) -> 2√8 -> 4√2 which would work and can be defined as (a ∈ R) because both a negative or positive can replace (a). But it would not work if the exponent in the radicand was odd because the end result of the radicand would be odd and that is not allowed. Overall, the end result of the radicand before must be positive (greater or equal to 0) if the index is even for it to be defined, and if the index is odd then the variable is equal to all real numbers.