Advanced math question! HELP!?

Question

ok so im working with factoring trinomials and i kno how to do it but i got confused when they incorporated exponents intothe middle part sooo for example:

x^4 - 16x^2y^2 + 55y^4

and could u please explain so i can do the other ones on my own??? thank u soooo much!!!

oh and another thing i was struggling with was slope:

#23. the vertices of a triangle are A (2,3) B (5,-2) and C (-1,-4) Find the slope of each side of hte triangle ABC. i think i have an idea tho....is it x sub 1 - x sub 2 over y sub 1 - y sub 2?? and would u just use the cordinates that make up that specific line?

aonce again THANK YOU SOO MUCH TO EVERYONE THAT HELPS!!! GREATLY APRECEATED!!!

Answer 1

x^4 - 16x^2y^2 + 55y^4

Let x^2 = a  & y^2 = b - then

a^2 - 16ab + 55 b^2

Next find two numbers which when multiplied give + 55 and added give -16

To find the two numbers, as a side sum write down all the factors of '55'

55 x 1 = 55

11 x 5 = 55

This factors to

(a - 11b)(a - 5b)

Substituting back for 'x' & 'y'

(x^2 - 11y^2)(x^2 - 5y^2)

These two factors will further factorise  as they both contain 'squared' terms.

(x - sqrt(11)y)(x + sqrt(11)y)(x - sqrt(5)y)(x + sqrt(5)y)

Fully factorised!!!!

Taking any pair of coordindates.

To find the slope between A & B

Then use the coordinates of 'A' & 'B'

A = ( 2,3)  & B = ( 5,-2)

Using these values

slope(AB) = (3 --2) / (2- 5) = 5/-3 = -5/3 = -1 2/3

NB

To find the slopes from the coordinates of any two points use

(y(1) - y(2)) / (x(1) - x(2))

Note the 'y' values go on the top and the 'x' values on the bottom.

The values a substracted so be careful of double negatives - they become positive.

The 'x' & 'y' coordinates of the first term must be in the left hand position.  

Similarly for the slope of BC and similarly again for the slope of AC

Answer 2

For the 1st one, the answer is (x^2-11y^2)(x^2-5y^2)

This is so because for the second terms (y^2), their sum (-11 + -5) equals the middle coefficient, while their product (-11 x -5) equals the final coefficient.

As for the 2nd problem, the slope is the other way around, y sub 1-y sub 2 over x sub 1 - x sub 2.

The answers are as follows:

AB:  3- (-2) /2-5 = -5/3

BC:  -2 - (-4)/5- (-1) = 1/3

AC:  3- (-4)/2- (-1) = 7/3

Answer 3

1) X^4-16X^2Y^2+55Y^4=

(X^2-11Y^2)(X^2-5Y^2) (by inspection)

The expression given has no problem!

The exponential expressions of the three terms match,

x^4,x^2y^2, y^4 are all terms of order 4.(note: 2+2=4 in

the middle term)

2) Two points can determine a straight line,hence its slope.

So, the slope of AB=(-2-3)/(5-2)=-5/3,

the slope of BC=(-4-(-2))/(-1-5)=1/3,

the slope of CA=(-4-3)/(-1-2)=7/3.

To find the equation of the line, use the formula:

y-b=m(x-a) known as the point-slope form,take

AB as an example, A(2,3) is on the line,then

y-3=(-5/3)(x-2)=>

5x+3y=19 is the required line.

Alright, you work out the rest yourself.

Answer 4

x^4 - 16x^2y^2 + 55y^4

= (x^2 - 11y^2)(x^2 - 5y^2

Slope of AB is (-2-3)/(5-2) = -5/3

Slope of BC is (-4-(-2))/(-1-5) = -2/-6 = 1/3

slope = difference in y values divided by difference in x values

Answer 5

I think you got confused because of the exponent 4 they introduced in the trinomial. So, let's begin by factoring this familiar trinomial:

a^2 - 16ab + 55b^2

The first 2 terms a^2 + 2*a*8b of the polynomial are also the first 2 terms of the expansion of the square of the difference (a - 8b). So we can complete the square by adding the third term (8b)^2

a^2 - 16ab + 55b^2 = a^2 - 2a*8b + (8b)^2 - (8b)^2 +55b^2

a^2 - 16ab + 55b^2 = (a - 8b)^2 - 64b^2 + 55b^2

a^2 - 16ab + 55b^2 = (a - 8b)^2 - 9b^2 = (a - 8b +3b)(a - 8b - 3b)

a^2 - 16ab + 55b^2 = (a - 5b)(a - 11b).

Above is what you have known how to do. Now, to factorize the polynomial x^4 - 16x^2y^2 + 55y^4, just think of x^4 as (x^2)^2, y^4 as (y^2)^2. x^2 and y^2 play the role of a and b in the above solved example.

x^4 - 16 x^2 y^2 + 55 y^4 = x^4 - 16 x^2 y^2 + 64 y^4 - 64 y^4 +55 y^4

= (x^2 - 8 y^2)^2 - 9 y^4 = (x^2 - 8 y^2)^2 - (3 y^2)^2

= (x^2 - 8 y^2 + 3y^2)(x^2 - 8 y^2 - 3y^2)

= (x^2 - 5 y^2)(x^2 - 11y^2)

= (x - y√5)(x + y√5 )(x - y√11)(x + y√11)

As for evaluating the slopes of AB, BC, CA, you are right in calculating the difference of the x-coordinates, of the y-coordinates then make the ratio. Only, you put things upside down. Remember this: slope = Δy / Δx  

For example: Evaluate the slope of AB

Δx = x(B) - x(A) = 5 - 2 = 3

Δy = y(B) - y(A) =-2 - 3 = -5

slope(AB) = [y(B) - y(A)] / [x(B) - x(A)] = -5/3

Answer 6

(x^2-11y^2)(x^2-5y^4)

Answer 7

Hey I dunno bout the first part but I think for the slope it could be y2 -y1 over x2 - x1