having trouble understanding 2 math problems.?

Question

Factor completely
1. x^5y^3-3x^4y^2-28x^3y



2. x^3-2x^2+6x-12

Answer 1

Hi,

a) Factor out a GCF of x^3y. This becomes:

x^3y(x^2y^2-3xy-28)

The trinomial can be factored into 2 binomials starting with xy, one with a minus, one with a plus. 28 splits into 7 and 4.

x^3y(xy-7)(xy+4)

That's your answer.

b) No GCF this time. With 4 terms look at 2 terms at a time for a GCF. First 2 have a GCF of x^2 and the last 2 have a GCF of 6.
x^3-2x^2+6x-12
x^2(x-2)+6(x-2)
Outsides go together as one factor. The other is the common factor.
(x^2+6)(x-2)

Those are your factors.

I hope that helps.

Just a note... If the problem had been

x^3-2x^2-8x+16, you'd still factor x^2 out of the first 2
x^2(x -2) but then to get the SAME thing left behind inside the other parentheses, you don't pull out 8, you pull out a negative 8 so parentheses will match.

x^2(x -2)-8(x-2) Factors would then be
(x^2-8)(x-2)

Answer 2

That in simple terms ability there's a quickly line on a graph that has the equation y = 2x + 2 in the beginning. Then they supply you a factor (-a million, 3) on yet another line that is parallel to the single from the equation. Parallel ability it has the comparable slope so which you will start up by utilising finding at y = 2x + 2 returned. The 2x ability the slope is two. (The coefficient of x = slope) Now you recognize the slope of the line with the factor (-a million, 3) you need to use the formulation to get the equation for it. the many times used formulation you start up with to locate the equation is y - y1 = m(x - x1) x1 and y1 are the x and y contained in the factor which you have have been given been given and m is the slope. So placed them in... y - 3 = 2(x - -a million) or y - 3 = 2(x + a million) do away with the bracket y - 3 = 2x + 2 Get y on that is very own y = 2x + 2 + 3 y = 2x + 5 and that's the equation for the different line confident evidently messy yet as quickly as you bear in mind "Get the slope first then use the formulation" you're guffawing. Have a bypass on the 2nd then do some hundred greater :-)

Answer 3

1. y^3 x^5 - 3y^ 2 x^4 - 28y x^3
= x^3 y (x y - 7) (x y + 4)

2. x^3 - 2x^2 + 6x - 12
= +- (x-2) (x^2+6)

It wont be much help in the exam, but this site will solve these equations for free (use factoris)
http://wims.unice.fr/wims/wims.cgi?session=OIA159B1C2.1&+lang=en&+module=home&+cmd=new&+search_category=T

use the other's calculations to understand the working

Answer 4

1. x^5y^3-3x^4y^2-28x^3y
=x^3y(x^2y^2-3xy-28)
= x^39(xy-7)(xy+4)

2. x^3-2x^2+6x-12
= x^2(x-2) +6(x-2)
= (x^2+6)(x-2)

Answer 5

Are you sure you don't mean x^6y^3-...

Because in that case you can write w=x^2y and reduce the equation to w^3-3x^2-28w, factor out a w and then you have an easy quadratic.

On the second one I would factor by grouping:
x^2(x-2)+6(x-2), and then regroup.

Answer 6

1)x3y(x2y2-3xy-28)
x3y(xy -7)(xy+4)

2) (x-2)(x2 +6)

Answer 7

1. x^3y(x^2y^2-3xy-28)
x^2y^2-7xy+4xy-28
xy(xy-7)+4(xy-7)
xy = 7 or xy = -4

Answer 8

1.) 30x^-2(-y)

2.) 12-7x

hope i got 'em! im pretty good in my algebra class..