• Textual 7 1 4 X 4 1 4 Ceramic Tile
• Textual 7 1 4 X 4 X 2
• Textual 7 1 4 X 4

Learning Outcomes

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• Use the least common denominator to eliminate fractions from a linear equation before solving it
• Solve equations with fractions that require several steps

You may feel overwhelmed when you see fractions in an equation, so we are going to show a method to solve equations with fractions where you use the common denominator to eliminate the fractions from an equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions.

Pay attention to the fact that each term in the equation gets multiplied by the least common denominator. That’s what makes it equal to the original!

EXAMPLE

Solve: [latex]frac{1}{8}x+frac{1}{2}=frac{1}{4}[/latex].

Solution:

In the last example, the least common denominator was [latex]8[/latex]. Now it’s your turn to find an LCD, and clear the fractions before you solve these linear equations.

Notice that once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve!

Solve equations by clearing the Denominators

1. Find the least common denominator of all the fractions in the equation.
2. Multiply both sides of the equation by that LCD. This clears the fractions.
3. Isolate the variable terms on one side, and the constant terms on the other side.
4. Simplify both sides.
5. Use the multiplication or division property to make the coefficient on the variable equal to [latex]1[/latex].

Textual 7 1 4 X 4 1 4 Ceramic Tile

Here’s an example where you have three variable terms. After you clear fractions with the LCD, you will simplify the three variable terms, then isolate the variable.

Example

Solve: [latex]7=frac{1}{2}x+frac{3}{4}x-frac{2}{3}x[/latex].

Solution:
We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.

Now here’s a similar problem for you to try. Clear the fractions, simplify, then solve.

Caution!

One of the most common mistakes when you clear fractions is forgetting to multiply BOTH sides of the equation by the LCD. If your answer doesn’t check, make sure you have multiplied both sides of the equation by the LCD.

In the next example, we’ll have variables and fractions on both sides of the equation. After you clear the fractions using the LCD, you will see that this equation is similar to ones with variables on both sides that we solved previously. Remember to choose a variable side and a constant side to help you organize your work.

Example

Solve: [latex]x+frac{1}{3}=frac{1}{6}x-frac{1}{2}[/latex].

Solution:

Now you can try solving an equation with fractions that has variables on both sides of the equal sign. The answer may be a fraction.

In the following video we show another example of how to solve an equation that contains fractions and variables on both sides of the equal sign.

In the next example, we start with an equation where the variable term is locked up in some parentheses and multiplied by a fraction. You can clear the fraction, or if you use the distributive property it will eliminate the fraction. Can you see why?

EXAMPLE

Solve: [latex]1=frac{1}{2}left(4x+2right)[/latex].

Solution:

Now you can try solving an equation that has the variable term in parentheses that are multiplied by a fraction.

Purplemath

'The weird case' of quadratic factoring is where it doesn't seem like we're factoring a quadratic, but we kind-of are. We need to be clever with these, but they reduce to little more than pattern-recognition, once you catch on to how to do them. The expressions are said to be 'quadratic in form', and we can apply the usual factorization methods to them.

• Factor x⁴ – 2x² – 8

At first glance, this does not appear to be a quadratic — and, in technical terms, it isn't. But this expression is quadratic in form, meaning that it can be restated as a quadratic, it follows the same patterns, and it can be factored by using the same techniques.

MathHelp.com

This polynomial has three terms (so it's a 'trinomial'). Taking a closer look at those exponents, I see that the power on the leading term is 4, and the power on the middle term is 2, which is half of 4. In a 'regular' quadratic, I would have the powers 2 and 1, where 1 is half of 2. In either case, the third term is just a number. So this polynomial follows the quadratic pattern; it is a quadratic 'in x²', rather than 'in x'. In fact, I can rewrite the expression as:

(x²)² – 2(x²) – 8

My leading coefficient is just (the 'understood') 1, so this is a case of simple quadratic factoring. I'll draw my parentheses as usual, but I'll put an x² at the beginning of each factor:

Factors of –8 that add to –2 are –4 and +2, so I get:

(x² – 4)(x² + 2)

Because this didn't start out as a regular quadratic, but instead as a polynomial in quadratic form, I still need to check to see if there is any further factoring possible. In this case, I still have a difference of squares, x² – 4, that I can factor. This gives me a final answer of:

Some books will encourage you to switch variables. Using the above example, they'd have you plugging, say, a 'y' in the original expression, in place of the 'x²', so you'd convert the original polynomial to:

y² – 2y – 8

After this substitution, the result is clearly a quadratic, which easily factors as:

Then you'd be expected to back-substitute the 'x²' for the 'y' to get:

(x² – 4)(x² + 2)

The entire process would look like this:

x⁴ – 2x² – 8

(x²)² – 2(x²) – 8

y² – 2y – 8

(y – 4)(y + 2)

(x² – 4)(x² + 2)

Many ('most'?) students find this 'substitution and back-substitution' process very confusing. I'm glad to be able to tell you that it is not at all necessary. If you're one that finds it helpful, more power to you. Use whatever works better for you.

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What follows are some fairly typical examples of weird quadratics, being non-quadratic polynomials which are of quadratic form. Note that there are always three terms in a quadratic-form expression, and the power (that is, the exponent) on the middle term is always half of the power on the leading term.

• Factor x⁶ + 6x³ + 5

This polynomial has three terms, and the degree of the middle term, being 3, is half of the degree of the leading term, being 6. So this is 'a quadratic in x³'; in fact, it's a simple quadratic to factor.

I need factors of +5 that add to +6, so I'll use +1 and +5 to split the middle term's coefficient:

I know that, because I didn't start with an actual quadratic, I might not actually be done with this factorization, so I look at the two factors I've got. I see that the second of these factors is a sum of cubes, which is something that I can factor further:

x³ + 1

(x)³ + (1)³

(x + 1)(x² – x + 1)

Putting this all together, my complete answer is:

In these two examples, after I'd factored the quadratic-form expression, I still had to do some more factoring. This will not always be the case, but will often be the case on tests, when the instructor will be checking to see if you're on top of your game. So keep in mind that, just because you've completed one factoring step, this doesn't mean that you've completed the exercise.

• Factor x^2/3 – x^1/3 – 6

This expression isn't even a polynomial, since polynomials are required to have whole-number exponents. However, this expression does have three terms, and the degree on the middle term is half of the degree on the leading term; and the third term is just a number. So this is in quadratic form; it's 'a quadratic in x^1/3'. And, because the leading coefficient is just 1, this is actually a simple case of factoring.

I need factors of –6 that add to –1. I'll use –3 and +2 to split the middle term's coefficient. This gives me:

x^2/3 – x^1/3 – 6

(x^1/3)² – 1(x^1/3) – 6

(x^1/3 – 3)(x^1/3 + 2)

There's nothing more that I can do with this, so I'm done. My answer is:

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• Factor x + 5x^1/2 + 4; express your answer in radical form.

This expression isn't entirely a polynomial, because the one-half power on the middle term's variable. But the expression has three terms, and the power on the middle term is half of the power on the leading term (being the understood 1), so this is of quadratic form; it's a quadratic in x^1/2, and it's a simple case of factoring.

x + 5x^1/2 + 4

(x^1/2)² + 5(x^1/2) + 4

(x^1/2 + 4)(x^1/2 + 1)

Because the one-half power is the same as the square root, my answer, 'in radical form', is:

(sqrt[x] + 4)(sqrt[x + 1)

You may find yourself presented with an expression to factor, where it's stated like this:

x + 5×sqrt[x] + 4

If so, then convert the square root to a one-half power, and factor as usual. Then convert back to radical form, because they generally (unless specified otherwise) expect and require the answer to be in the same format as was the original question.

• Factor 4x⁴ – 25

This has only two terms, and nothing comes out of both. So they've thrown me a curveball; this is actually just a difference of squares. I will factor in the usual way:

Checking, I see that neither of these factors is a difference of squares, nor a sum or difference of cubes, so there's nothing more I can do. My answer is:

(2x² – 5)(2x² + 5)

• Factor (x – 3)⁴ + 2(x – 3)² – 8, and simplify.

This expression is a polynomial, and it does have three terms, though they're big lumpy ones. The power on the middle term (namely, 2) is half of the power on the leading term (namely, 4), so this is quadratic in form; it's a quadratic in (x – 3)². I will factor as usual....

Well, actually, because I'm lazy and this particular expression has such big lumpy terms, I'll do the substitution thing. But I won't bother with a variable; I'll just use a shape.

(x – 3)⁴ + 2(x – 3)² – 8

▭² + 2▭² – 8

(▭ + 4)(▭ – 2)

((x – 3)² + 4)((x – 3)² – 2)

I'm not done yet, because I've got a squared binomial in there that I can multiply out, and then simplify inside each of the two resulting parentheticals:

Textual 7 1 4 X 4 X 2

((x² – 6x + 9) + 4)((x² – 6x + 9) – 2)

(x² – 6x + 13)(x² – 6x + 7)

Now I need to check to see if either of these simple-case quadratics factors further.

In the first parenthetical, I need factors of +13 that add to –6; the only factor pair for 13 is 1 and 13, so the first quadratic doesn't factor. In the second parenthetical, I need factors of +7 that add to –6; the only factor pair for 7 is 1 and 7. But I'm multiplying to a 'plus', so the factors have to have the same sign; neither –1 – 7 nor +1 + 7 equals –6, so this quadratic doesn't factor, either.

My final answer then is:

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I can check my answer this last exercise by multiplying the two polynomial factors in my answer, multiplying everything out in the original expression, and verifying (by looking at them) that they both simplify to the same fourth-degree polynomial:

x⁴ – 12x³ + 56x² – 120x + 91

Textual 7 1 4 X 4

In fact, you can use this method to check any factorization answer. If you multiply everything back together, and you get what you'd started with, you then know that your factorization was correct. This can be handy, if you've got some time left over after you've finished answering everything on the test.

However, if I'd completely multiplied out the original expression, I may not have been able to figure out the factorization. Even though the original polynomial was lumpen, it paid off to work with it as it stood, rather than trying to simplify too early.

You can use the Mathway widget below to practice factoring expression that are quadratic in form. Try the entered exercise, or type in your own exercise. Then click the button to compare your answer to Mathway's.

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URL: https://www.purplemath.com/modules/factquad4.htm