Factor b^2 - ab - 6a^2, y^3 - 12^2 + 36y completely

[Debbiesmith]

Help me to factor completely: b^2 - ab - 6 a^2 and help me with factoring completely: y^3 - 12^2 + 36y

Also, help with factoring by grouping x^3 - 5x^2 + 2x - 10 ... :shock:

I will need the help by tomorrow!

 

[Loren]

factor completely: b^2 - ab - 6 a^2

Fill in the blanks to make these true statements.

(b + ___)(b - ___) = b^2 - ab - 6 a^2


y^3 - 12^2 + 36y (typo)???
y^3 - 12y^2 + 36y = y(y^2 -12y + 36) = y(y - ___)(y - ___) = y(y - ___)^2


x^3 - 5x^2 + 2x - 10
\(\displaystyle x^2(x-5) + 2(x-5)\) I'll leave the rest to you.

 

[Denis]

Re: Intermediate Algebra

I will need the help by tomorrow!

Show your attempts next time; easier for us to help;
if you don't, you'll be lucky if you get help by next year :wink:

 

[arthur ohlsten]

b^2-ab-6a^2

factors of b^2= b,b
factors of 6a^2= 6a,a..2a,3a
we want a set whose difference of cross products =-ab
b,b and 2a,3a does it
[b-3a][b+2a] answer

there is a possible error with the second problem because the factors are not integers. instructors usually assign problems with integer solutions

x^3-5x^2+2x-10 rearrange by odd powers of x and even powers of x
[x^3+2x]+ [-5x^2-10x^0] factor a x from 1st and -5 from 2nd term
x[x^2+2]-5[x^2+2] factor [x^2+2]
[x^2+2][x-5] answer

I assume you do not want to factor x^2+2=[x+i sqrt2][x-isqrt2]

Arthur

 

[stapel]

Help me to factor completely...
Also, help with factoring by grouping...

Since you were unable to get started, I will guess that you missed the weeks when this was covered in class, and are needing lessons...? :?:

. . . . .Google results for "simple factoring"

. . . . .Google results for "factoring quadratics"

The first link will help you with the "by grouping" problem, and with the first step (taking the common factor out front) in the second "factoring completely" problem. The second link will help you with both of the "factoring completely" problems. :wink:

Eliz.