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- YanaSidlinskiy

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- schrodinger

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- YanaSidlinskiy

1.) Factor out the GCF. \[16x^7-5x^5+5x^4\]

- YanaSidlinskiy

My answer: \[x^4(16x^3-5x+5)\]

- YanaSidlinskiy

2.) Factor out the GCF:\[60x^2-12xy+28x\]
My answer:\[4x(15x-3y+7)\]

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- YanaSidlinskiy

3.) Factor by grouping. \[4x^2-8xy-3x+6y\]
My answer: \[(4x-3)(x-2y)\]

- YanaSidlinskiy

4.) Factor:\[x^2-x-42\]
My answer:\[-6,7\]
>>>> I'm thinking I have this down wrong or something.. <<<<

- Nnesha

you don't need to solve for x ^^^ -6 and 7 are zero (x-intercept )

- anonymous

4.) Is your answer the zeros?

- YanaSidlinskiy

5.) Which of the following trinomials has a greatest common factor that can be factored out first? \[100x^2+140x+49\]
\[9x^2+24x+16\]
\[x^2+4x+4\]
\[8x^2+16x+8\]

- YanaSidlinskiy

My answer is the last one and @mathway no. My answer for number for as you can see above...

- YanaSidlinskiy

6.) Factor completely.
\[1-4x^2\]
My answer: \[(1-2x)^2\]

- anonymous

Then you're wrong. (x-6)(x+7) is not equal to the polynomial. You might want to check your signs.

- YanaSidlinskiy

Ok, thought so too!

- Nnesha

6)hint apply difference of squares method

- YanaSidlinskiy

6.) Isn't that what I did??

- Nnesha

difference of squares \[\huge\rm a^2-b^2 =(a\color{Red}{-}b)(a\color{reD}{+}b)\] one parentheses with negative sign and other one should be positive

- Nnesha

\[\huge\rm -4x^2+1\] first take out the negative sign
\[\huge\rm -(4x^2-1)\]

- YanaSidlinskiy

So it'd be: \[2x^2-1\]?

- Nnesha

that is one of the factor
(a-b)

- YanaSidlinskiy

Ohh! Wait. I know what I did wrong.. It'd be: \[(1-2x)(1+2x)\]

- YanaSidlinskiy

Is that right? John I see you spying ;)

- Nnesha

well that's right
or you can leave it as \[-(2x-1)(2x+1)\]
remember we already factor out the negative sign :=)

- Nnesha

thanks o^_^o

- YanaSidlinskiy

So the one is a negative then?

- Nnesha

i would leave it as -(2x-1)(2x+1)
which is same as (-2x+1)(2x+1)
|dw:1438901505115:dw|

- YanaSidlinskiy

Ahh, ok. What about number 4? What's up with it??

- YanaSidlinskiy

Would it be (x+6)(x-7)?

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @YanaSidlinskiy
6.) Factor completely.
\[1-4x^2\]
My answer: \[(1-2x)^2\]
\(\color{blue}{\text{End of Quote}}\)
(a+b)(a+d) <-- these are factors
you don't have to solve for x
(x+6)(x-7) are factors
x=-6,7 are zeros(x-intercept ,solutions )

- Nnesha

yes that's right

- johnweldon1993

\[\large x^2 -x - 42=0\]
We have -7 and 6 as zeros...meaning
\[\large (x+7)(x-6)=0\]
right?

- YanaSidlinskiy

Yeah.

- johnweldon1993

Make sure if you mean "yeah" or not...because we have 2 different answers :P

- johnweldon1993

And that was a test :P
You have it correct @YanaSidlinskiy

- YanaSidlinskiy

Ok, well for my next question....
7.) Solve by factoring. \[b^2=32-4b\]
My answer: \[-8,4\]
OMG @johnweldon1993 I'M CALLING YOU ON THAT ONE!!!!! After all the struggle...... YOU JUST HAD TOO!!!!

- johnweldon1993

I was trying to test you because you've been doing so good!!
You are correct
The REAL zeros are
-6 and 7
So indeed
\[\large (x+6)(x-7)\] is correct

- johnweldon1993

Lol you're right, I want you confident with your answers!

- YanaSidlinskiy

LOL. I was about to go off and say that -6+7 is not -1 lol.. Omg that was a heartbeat. Say sorry hhaa;) Jkjk.. Anyways..

- YanaSidlinskiy

I've posted... #7..... Is it right?

- YanaSidlinskiy

PLEASE SAY YES haha.

- Nnesha

\(\huge\color{green}{\checkmark}\)

- johnweldon1993

Oh sorry...looked away...indeed!

- YanaSidlinskiy

Why would you look away when it's me... SUNSHINE hahha;;) Jkjk.... Anyways..
9.) \[x^2-4x+4=0\]
My answer: The equation has no real solutions..

- YanaSidlinskiy

^ I was supposed to be solving that by the quadratic formula..

- johnweldon1993

Hmm...are you sure?
You didnt get ANY zeros?

- YanaSidlinskiy

Is this a test John?

- johnweldon1993

Lol no, a legitimate question...lets do the formula out
\[\large \frac{4 \pm \sqrt{16 - 4(1)(4)}}{2}\]

- YanaSidlinskiy

\[\frac{ 4\pm \sqrt{16-16} }{ 2 }\]
I tried ignoring the "-"..... Lol. Even if this is right, it'd factor out 16-16=0..

- YanaSidlinskiy

Ohh... great... Here comes the PRO.

- johnweldon1993

No that is correct
So if we have
\[\large \frac{4 \pm \sqrt{0}}{2}\]
Whats the square root of 0?

- YanaSidlinskiy

0..

- johnweldon1993

Right, so it is 4 plus or minus 0 (which is just 4 in either case) divided by 2
\[\large \frac{4}{2} = ?\]

- YanaSidlinskiy

2.

- johnweldon1993

And that is your zero :)

- YanaSidlinskiy

So, wait... x=0?

- YanaSidlinskiy

I mEAN 2=X? Hhaa.

- YanaSidlinskiy

Or more like x=2 lol...

- johnweldon1993

There you go :)
You can always check all these too...plug in 2 into your equation for 'x' and see if it comes out right

- YanaSidlinskiy

10.) If S=-16t^2+64t represents the height of a rocket, in feet, t seconds after it was fired, when will the rocket hit the ground?
I got 4 seconds...

- YanaSidlinskiy

Lol, okie.

- johnweldon1993

Ehh instead of going into a whole thing...yes correct lol

- YanaSidlinskiy

Lol, okie. Ridiculous one for the last haha;) So I'm supposed to be solving this equation using the quadratic formula... I do not know like what to do.. I mean I do but. Idk lol.. \[y^2=3y+18\]

- YanaSidlinskiy

I can do this, right? \[y^2-3y-18 = 0\]

- johnweldon1993

Its the same thing...just using 'y' instead of 'x'
But same process
And yes...that is exactly what you will do!

- YanaSidlinskiy

Ok, you can have a break while I solve this... And tell me if I'm right..

- johnweldon1993

"Makes a sandwich" :P

- YanaSidlinskiy

LOL.

- YanaSidlinskiy

6?

- johnweldon1993

...and... remember the quadratic formula always gives you 2 answers

- YanaSidlinskiy

(6,-6)

- YanaSidlinskiy

*-3

- johnweldon1993

Not quite...
\[\large \frac{3\pm \sqrt{9 - 4(1)(-18)}}{2}\]
\[\large \frac{3\pm \sqrt{9 + 72}}{2}\]
\[\large \frac{3 \pm \sqrt{81}}{2}\]
Ahh there you go...you got it
The answers are 6 and -3

- YanaSidlinskiy

So, (6,-3)?

- johnweldon1993

Just dont write it like that...it is not a coordinate
Just y = 6 and y = -3 are solutions to this equation

- YanaSidlinskiy

|dw:1438903906614:dw|

- johnweldon1993

Lol <3

- YanaSidlinskiy

Okie, thanks again. I gotta go cuz college closes in 30 minutes and that's where I'm at haha. So I gotta go home rn..

- johnweldon1993

Of course :) have fun!

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