factoring by grouping

[PreAlgebraDummy15]

I am now in algebra! I need some help with these:

1. 27y (the y is squared)-48 y (the y is to the 4th power)
2. 6z (z is to the 3rd power) + 3z (z is squared) + 2z + 1
3. -64 + c (c is to the 14th power)
4. 8a (a is to the 3rd power)-12a (a is to the second power + 6a-9

If anyone could please help that would be great thanks!

 

[stapel]

1. 27y (the y is squared)-48 y (the y is to the 4th power)
2. 6z (z is to the 3rd power) + 3z (z is squared) + 2z + 1
3. -64 + c (c is to the 14th power)
4. 8a (a is to the 3rd power)-12a (a is to the second power + 6a-9

Please correct or confirm:

. . . . .1) 27y² - 48y⁴

. . . . .2) 6z³ + 3z² + 2z + 1

. . . . .3) -64 + c¹⁴

. . . . .4) 8a³ - 12a² + 6a - 9

Thank you.

Eliz.

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[PreAlgebraDummy15]

if <sup> means squared then that is right

 

[stapel]

if <sup> means squared then that is right

Actually, "sup" means "superscript"; in this case, it raises numbers above the baseline, indicating exponents.

1) This starts as just simple factoring. Take the common factor out front. (That is, this is of the form "6x + 8 = 2(3x + 4)".) Then apply the difference-of-squares formula you've memorized to what's left inside the parentheses.

2) Factor "in pairs". (That is, this is of the form "4x^3 - 8x^2 - 5x + 10 = 4x^2(x - 2) - 5(x - 2) = (x - 2)(4x^2 - 5)".)

3) Rearrange as c^14 - 64 = (c^7)^2 - 8^2, and apply the appropriate formula.

4) This works like (2).

If you get stuck, please reply showing all the steps you have tried. Thank you.

Eliz.

Edit: Changing superscript notation to something the student's browser should be able to parse.

 

[PreAlgebraDummy15]

27y² - 48y⁴

would it be:

3 (9y¹-16y²

 

[stapel]

27y² - 48y⁴ would it be:

3 (9y¹-16y²

What happened to the exponents? For instance, how did "27y^2" become "3(9y)"?

Thank you.

Eliz.

 

[PreAlgebraDummy15]

9 y ²

would it be that?

 

[stapel]

9 y ²
would it be that?

Yes. Now make the appropriate correction to the other term, and factor out the common variable factor (that is, put it out front with the "3"). Then apply the difference-of-squares formula to what's left.

Eliz.

 

[PreAlgebraDummy15]

is the final answer:

3 (9y²-16y⁴)

3 (9y²+16y⁴

 

[PreAlgebraDummy15]

ok I'm working on this one, am I finished?:

8a³ -12a² + 6a-9

(8a³-12a² +6a-9

4a (2a ²-3a1) + 3 (2a-3)

 

[PreAlgebraDummy15]

More factoring by grouping help!

5x⁴ -80

5(1x³-16) 5(1x³+16)


Is that the right answer? I factored 5 & 80 by 5 (5 goes into 5 once and 5 goes into 80 16 times. ) Thanks!

 

[PreAlgebraDummy15]

NEED HELP! PLEASE, SUPER URGENT! THANKS!

ok i would really appreciate help at this point!! Please I really need someone, thanks.

 

[Denis]

Re: More factoring by grouping help!

PLEASE show 5 times x to the power 4 this way: 5x^4 (^ means "to the power").

For all the good it'll do you:
5x^4 - 80
= 5(x^4 - 16)
= 5(x^2 - 4)(x^2 + 4)
= 5(x - 2)(x + 2)(x^2 + 4)

Notes:
x^a times x^b = x^(a + b)

a^2 - b^2 = (a - b)(a + b)

 

[PreAlgebraDummy15]

factoring by grouping for Denis or anyone

thanks so much now how would u do this one:

1-y^8

and this one:

625-m^4

would it be: 5(125-m^4)?

Also:

6z^3+3z^2+2z+1

thank-you

 

[Denis]

Re: factoring by grouping for Denis or anyone

and this one:
625-m^4
would it be: 5(125-m^4)?

5(125 - m^4) = 625 - 5m^4

I'm not doing any more of your homework;
you need to have a serious talk with your teacher.

 

[stapel]

NEED HELP! PLEASE, SUPER URGENT! ok i would really appreciate help at this point!! Please I really need someone

Please note that there is no paid staff waiting on-hand to provide instant replies. The tutors are volunteers who surf by when they are able. You should be prepared to wait hours, if not days, for an online response.

now how would u do this one...

Please post new questions as new threads, not as replies to old threads, where they tend to be overlooked.

And please start showing some work, especially when the new exercises you're posting are identical (in form) to those already completely explained and worked out for you.

Thank you.

Eliz.