Rational Expressions Practice/Guess and Check With Quadratics

[spaceshowfeature1]

Where can I find practice for more problems like these: 5x^3-17x^2-12x/x^2-4x, and 6x^2+13x+5/3x^2+26x+35. The first problems involves canceling the x's as coefficients. ex:x(5x+3)/x=5x+3. The second problem involves lots of guessing and checking when factoring the two quadratics. How do I factor a quadratic like this, without guess and check? Where can I find more problems like these? I have redone the problems I got wrong. (these ones) Is that enough? After I get a problem wrong, and redo it I feel obligated to do practice problems like the ones I got wrong. Do I need to do more problems? Thank you for your input!

 

[stapel]

Where can I find practice for more problems like these?

(a) 5x^3-17x^2-12x/x^2-4x

(b) 6x^2+13x+5/3x^2+26x+35

As posted, the expressions mean the following:

. . . . .\(\displaystyle \mbox{(a) }\, 5x^3\, -\, 17x^2\, -\, \dfrac{12x}{x^2}\, -\, 4x\)

. . . . .\(\displaystyle \mbox{(b) }\, 6x^2\, +\, 13x\, +\, \dfrac{5}{3x^2}\, +\, 26x\, +\, 35\)

However, from your subject line, I suspect that the grouping symbols and instructions were omitted, and that the intended exercises are:

. . . . .\(\displaystyle \mbox{(a') Simplify }\, \dfrac{5x^3\, -\, 17x^2\, -\, 12x}{x^2\, -\, 4x}\)

. . . . .\(\displaystyle \mbox{(b') Simplify }\, \dfrac{6x^2\, +\, 13x\, +\, 5}{3x^2\, +\, 26x\, +\, 35}\)

The first problems involves canceling the x's as coefficients; for example: [x(5x+3)]/x=5x+3.

So you factored out the \(\displaystyle x\) from each of the numerator and denominator:

. . . . .\(\displaystyle \mbox{(a') }\, \dfrac{5x^3\, -\, 17x^2\, -\, 12x}{x^2\, -\, 4x}\, =\, \dfrac{x\, (5x^2\, -\, 17x\, -\, 12)}{x\, (x\, -\, 4)}\)

Making sure to note that \(\displaystyle x\) cannot equal 0, you cancelled to get:

. . . . .\(\displaystyle \mbox{(a') }\, \dfrac{5x^2\, -\, 17x\, -\, 12}{x\, -\, 4}\)

However, this is not the complete simplification. Now the quadratic in the numerator must be factored or, which ends up in the same place, the numerator must be divided by the denominator. Then cancel the common factor (noting the restriction on the domain) to get the final form.

The second problem involves lots of guessing and checking when factoring the two quadratics.

Did your class not cover how to factor quadratics? To learn the usual method for this, try here.

Once you have studied the lesson and learned the basic terms and techniques, apply them to the two quadratics in the second exercise. For the numerator, start by finding factors of (6)(+5) = +30 that add to +13. For the denominator, start by finding factors of (3)(+35) = +105 that add to +26.

 

[mmm4444bot]

Where can I find practice for more problems like these: (5x^3-17x^2-12x)/(x^2-4x) and (6x^2+13x+5)/(3x^2+26x+35) …. How do I factor a quadratic like [these numerators and denominators] …

There's a better method than guess-and-check. It's called 'factor by grouping', and it begins by multiplying the first and last coeffcients (A∙C)

For lessons and examples, google keywords factor by grouping using ac method

For practice, google keywords ac method factoring "practice problems"

 

[spaceshowfeature1]

As posted, the expressions mean the following:

. . . . .\(\displaystyle \mbox{(a) }\, 5x^3\, -\, 17x^2\, -\, \dfrac{12x}{x^2}\, -\, 4x\)

. . . . .\(\displaystyle \mbox{(b) }\, 6x^2\, +\, 13x\, +\, \dfrac{5}{3x^2}\, +\, 26x\, +\, 35\)

However, from your subject line, I suspect that the grouping symbols and instructions were omitted, and that the intended exercises are:

. . . . .\(\displaystyle \mbox{(a') Simplify }\, \dfrac{5x^3\, -\, 17x^2\, -\, 12x}{x^2\, -\, 4x}\)

. . . . .\(\displaystyle \mbox{(b') Simplify }\, \dfrac{6x^2\, +\, 13x\, +\, 5}{3x^2\, +\, 26x\, +\, 35}\)


So you factored out the \(\displaystyle x\) from each of the numerator and denominator:

. . . . .\(\displaystyle \mbox{(a') }\, \dfrac{5x^3\, -\, 17x^2\, -\, 12x}{x^2\, -\, 4x}\, =\, \dfrac{x\, (5x^2\, -\, 17x\, -\, 12)}{x\, (x\, -\, 4)}\)

Making sure to note that \(\displaystyle x\) cannot equal 0, you cancelled to get:

. . . . .\(\displaystyle \mbox{(a') }\, \dfrac{5x^2\, -\, 17x\, -\, 12}{x\, -\, 4}\)

However, this is not the complete simplification. Now the quadratic in the numerator must be factored or, which ends up in the same place, the numerator must be divided by the denominator. Then cancel the common factor (noting the restriction on the domain) to get the final form.


Did your class not cover how to factor quadratics? To learn the usual method for this, try here.

Once you have studied the lesson and learned the basic terms and techniques, apply them to the two quadratics in the second exercise. For the numerator, start by finding factors of (6)(+5) = +30 that add to +13. For the denominator, start by finding factors of (3)(+35) = +105 that add to +26.

lol I'm talking Algebra, and I just started school, but I'm very impatient and I'm teaching myself some Algebra. Anyway, thank you! This was very helpful!

 

[spaceshowfeature1]

There's a better method than guess-and-check. It's called 'factor by grouping', and it begins by multiplying the first and last coeffcients (A∙C)

For lessons and examples, google keywords factor by grouping using ac method

For practice, google keywords ac method factoring worksheet practice

Thank You mmm4444bot, Very Cool!

 

[Steven G]

Where can I find practice for more problems like these: 5x^3-17x^2-12x/x^2-4x, and 6x^2+13x+5/3x^2+26x+35. The first problems involves canceling the x's as coefficients. ex:x(5x+3)/x=5x+3. The second problem involves lots of guessing and checking when factoring the two quadratics. How do I factor a quadratic like this, without guess and check? Where can I find more problems like these? I have redone the problems I got wrong. (these ones) Is that enough? After I get a problem wrong, and redo it I feel obligated to do practice problems like the ones I got wrong. Do I need to do more problems? Thank you for your input!

Guess and check is the easiest and fastest ways to do these problems. You can always use the quadratic formula to help in factoring and this method will always work without guessing (but is almost always longer). Do you need to do more problems? My answer to that is to do enough problems until you start to get them all correct and they all start to look the same.