You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
help me !!!
- Thread starter denise
- Start date
G
Guest
Guest
x^1/2 is (x^1) / 2. I think you mean x^(1/2).
x^(1/2) - 3x
= x^(1/2) - 3(x^(1/2))^2
= x^(1/2)(1 - 3x^(1/2))
2(x + 3)^(-4/3) + (x + 3)^(-1/3)
= 2(x + 3)^(-4/3) + (x + 3)^(3/3)(x + 3)^(-4/3)
= 2(x + 3)^(-4/3) + (x + 3)(x + 3)^(-4/3)
= (2 + (x + 3))(x + 3)^(-4/3)
= (x + 5)(x + 3)^(-4/3)
x^(1/2) - 3x
= x^(1/2) - 3(x^(1/2))^2
= x^(1/2)(1 - 3x^(1/2))
2(x + 3)^(-4/3) + (x + 3)^(-1/3)
= 2(x + 3)^(-4/3) + (x + 3)^(3/3)(x + 3)^(-4/3)
= 2(x + 3)^(-4/3) + (x + 3)(x + 3)^(-4/3)
= (2 + (x + 3))(x + 3)^(-4/3)
= (x + 5)(x + 3)^(-4/3)
Hello, denise!
Just HOW do we factor: . x<sup>5</sup> + 3x<sup>2</sup> ?
First, we note that there are x's common to both terms.
Next, we use the <u>smaller</u> exponent to factor, right?
. . . In this case, "2" . . . So we factor out x<sup>2</sup>.
Then we write x<sup>2</sup> "out front"
. . . and in parentheses, we write "what's left".
How do we know "what's left" ?
. . . We are <u>dividing</u> out the x<sup>2</sup> . . . so we <u>subtract</u> exponents.
So, "inside", we have: . x<sup>5-2</sup> + 3x<sup>2-2</sup> . = . x<sup>3</sup> + 3x<sup>0</sup> . = . x<sup>3</sup> + 3
. . And <u>that's</u> why the answer is: . x<sup>2</sup>(x<sup>3</sup> + 3)
(Of course, we do all that without thinking all those steps, right?)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Let's try #2: . 2(x + 3)<sup>-4/3</sup> + (x + 3)<sup>-1/3</sup>
First, we note that there are (x + 3)'s common to both terms.
. . . How many should we take out?
We use the <u>smaller</u> exponent, remember?
. . . In this case, it is -4/3. . (Careful!)
We write x<sup>-4/3</sup> out front . . . Now, what's left?
We <u>subtract</u> exponents . . .
. . . For the first term, the exponent is: . (-4/3) - (-4/3) .= .0
. . . For the second term it is: . (-1/3) - (-4/3) .= .1
Hence, we have: . 2(x+3)<sup>0</sup> + (x + 3)<sup>1</sup> . = . 2·1 + x + 3 . = . x + 5
. . Therefore: . (x + 3)<sup>-4/3</sup>(x + 5)
It's not too bad once you see the <u>reasoning</u> behind it.
. . . And some practice wouldn't hurt, would it?
I usually have to baby-step my way through this type.i am having a lil trouble remembering how to factor. I'm not surprised; these are tricky!
1) x<sup>1/2</sup> - 3x
2) 2(x+3)<sup>-4/3</sup> + (x+3)<sup>-1/3</sup>
.
Just HOW do we factor: . x<sup>5</sup> + 3x<sup>2</sup> ?
First, we note that there are x's common to both terms.
Next, we use the <u>smaller</u> exponent to factor, right?
. . . In this case, "2" . . . So we factor out x<sup>2</sup>.
Then we write x<sup>2</sup> "out front"
. . . and in parentheses, we write "what's left".
How do we know "what's left" ?
. . . We are <u>dividing</u> out the x<sup>2</sup> . . . so we <u>subtract</u> exponents.
So, "inside", we have: . x<sup>5-2</sup> + 3x<sup>2-2</sup> . = . x<sup>3</sup> + 3x<sup>0</sup> . = . x<sup>3</sup> + 3
. . And <u>that's</u> why the answer is: . x<sup>2</sup>(x<sup>3</sup> + 3)
(Of course, we do all that without thinking all those steps, right?)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Let's try #2: . 2(x + 3)<sup>-4/3</sup> + (x + 3)<sup>-1/3</sup>
First, we note that there are (x + 3)'s common to both terms.
. . . How many should we take out?
We use the <u>smaller</u> exponent, remember?
. . . In this case, it is -4/3. . (Careful!)
We write x<sup>-4/3</sup> out front . . . Now, what's left?
We <u>subtract</u> exponents . . .
. . . For the first term, the exponent is: . (-4/3) - (-4/3) .= .0
. . . For the second term it is: . (-1/3) - (-4/3) .= .1
Hence, we have: . 2(x+3)<sup>0</sup> + (x + 3)<sup>1</sup> . = . 2·1 + x + 3 . = . x + 5
. . Therefore: . (x + 3)<sup>-4/3</sup>(x + 5)
It's not too bad once you see the <u>reasoning</u> behind it.
. . . And some practice wouldn't hurt, would it?