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1.) Factor out the GCF. \[16x^7-5x^5+5x^4\]
My answer: \[x^4(16x^3-5x+5)\]
2.) Factor out the GCF:\[60x^2-12xy+28x\] My answer:\[4x(15x-3y+7)\]

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3.) Factor by grouping. \[4x^2-8xy-3x+6y\] My answer: \[(4x-3)(x-2y)\]
4.) Factor:\[x^2-x-42\] My answer:\[-6,7\] >>>> I'm thinking I have this down wrong or something.. <<<<
you don't need to solve for x ^^^ -6 and 7 are zero (x-intercept )
4.) Is your answer the zeros?
5.) Which of the following trinomials has a greatest common factor that can be factored out first? \[100x^2+140x+49\] \[9x^2+24x+16\] \[x^2+4x+4\] \[8x^2+16x+8\]
My answer is the last one and @mathway no. My answer for number for as you can see above...
6.) Factor completely. \[1-4x^2\] My answer: \[(1-2x)^2\]
Then you're wrong. (x-6)(x+7) is not equal to the polynomial. You might want to check your signs.
Ok, thought so too!
6)hint apply difference of squares method
6.) Isn't that what I did??
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
\[\huge\rm -4x^2+1\] first take out the negative sign \[\huge\rm -(4x^2-1)\]
So it'd be: \[2x^2-1\]?
that is one of the factor (a-b)
Ohh! Wait. I know what I did wrong.. It'd be: \[(1-2x)(1+2x)\]
Is that right? John I see you spying ;)
well that's right or you can leave it as \[-(2x-1)(2x+1)\] remember we already factor out the negative sign :=)
thanks o^_^o
So the one is a negative then?
i would leave it as -(2x-1)(2x+1) which is same as (-2x+1)(2x+1) |dw:1438901505115:dw|
Ahh, ok. What about number 4? What's up with it??
Would it be (x+6)(x-7)?
\(\color{blue}{\text{Originally Posted by}}\) @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 )
yes that's right
\[\large x^2 -x - 42=0\] We have -7 and 6 as zeros...meaning \[\large (x+7)(x-6)=0\] right?
Yeah.
Make sure if you mean "yeah" or not...because we have 2 different answers :P
And that was a test :P You have it correct @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!!!!
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
Lol you're right, I want you confident with your answers!
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..
I've posted... #7..... Is it right?
PLEASE SAY YES haha.
\(\huge\color{green}{\checkmark}\)
Oh sorry...looked away...indeed!
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..
^ I was supposed to be solving that by the quadratic formula..
Hmm...are you sure? You didnt get ANY zeros?
Is this a test John?
Lol no, a legitimate question...lets do the formula out \[\large \frac{4 \pm \sqrt{16 - 4(1)(4)}}{2}\]
\[\frac{ 4\pm \sqrt{16-16} }{ 2 }\] I tried ignoring the "-"..... Lol. Even if this is right, it'd factor out 16-16=0..
Ohh... great... Here comes the PRO.
No that is correct So if we have \[\large \frac{4 \pm \sqrt{0}}{2}\] Whats the square root of 0?
0..
Right, so it is 4 plus or minus 0 (which is just 4 in either case) divided by 2 \[\large \frac{4}{2} = ?\]
2.
And that is your zero :)
So, wait... x=0?
I mEAN 2=X? Hhaa.
Or more like x=2 lol...
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
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...
Lol, okie.
Ehh instead of going into a whole thing...yes correct lol
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.. \[y^2=3y+18\]
I can do this, right? \[y^2-3y-18 = 0\]
Its the same thing...just using 'y' instead of 'x' But same process And yes...that is exactly what you will do!
Ok, you can have a break while I solve this... And tell me if I'm right..
"Makes a sandwich" :P
LOL.
6?
...and... remember the quadratic formula always gives you 2 answers
(6,-6)
*-3
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
So, (6,-3)?
Just dont write it like that...it is not a coordinate Just y = 6 and y = -3 are solutions to this equation
|dw:1438903906614:dw|
Lol <3
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..
Of course :) have fun!

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