Math genius needed I have to practice factoring and I am having a problem with this one (3y-2)^2-(y+1)^2?

Question

your help will be happily appreciated

Answer 1

(3y-2)^2-(y+1)^2
break this down in steps
(3y)^2 - 2(3y)(2) + 2^2 - (y^2 + 2(y)(1) +1^2)
9y^2 - 12y + 4 -y^2 - 2y -1
8y^2 - 14Y +3=0
Now you can solve the quadratic formula

ay^2 + by+ c = 0

y = [ -b ± sqrt(b^2 - 4ac) ] / 2a

Answer 2

James K is correct… but you wanted to factor this, didn’t you?

(3y-2)^2=9y^2-12y+4
(y+1)^2=y^2+2y+1
The difference is: 8y^2-14y+3

We're looking for an (ax + b)(cx + d) so that
ac = 8, bd=3, and (ad + bc)= -14

(b and d)
The factors of 3 are 1 and 3, or -1 and -3

(a and c)
The factors of 8 are 1 and 8, or -1 and -8, or 2 and 4, or -2 and -4

Now we need to get the combination of a,b,c, and d so that
ac+bd = -14
If you test several combinations you will eventually come up on
(4x – 1)(2x – 3)… I picked that combination because (-3)(4) + (-1)(2) = -12 – 2 = -14
Although (-4x + 1)(-2x + 3) also work… and for the same reason.

Answer 3

Factor this the same way you would factor a^2 - b^2, which factors as (a - b)(a + b).

The difference here is that you have whole binomials to treat as a difference of squares.

(3y - 2)^2 - (y + 1)^2

Note: Watch the bracketing when doing this!

This factors as

( [3y - 2] - [y + 1] ) ( [3y - 2] + [y + 1] )

Now, simplify each.

(3y - 2 - y - 1) (3y - 2 + y + 1)
(2y - 3) (4y - 1)

Answer 4

Use the special product "difference of squares":
a^2 - b^2 = (a + b)(a - b).

(3y - 2)^2 - (y + 1)^2

((3y - 2) + (y + 1))((3y - 2) - (y + 1))

(3y - 2 + y + 1)(3y - 2 - y - 1)

(4y - 1)(2y - 3)

Answer 5

(3y-2)^2-(y+1)^2

(3y-2)(3y-2)-(y+1)(y+1) F.O.I.L.

(9y^2-6y-6y+4)-(y^2+y+y+1) Use the Distributive Property

(9y^2-12y+4)-(y^2+2y+1) Add like terms

9y^2-12y+4-y^2-2y-1 Distribute (-)

8y^2-14y+3 Simplify

(4y-1)(2y-3) Factor

4y-1=0 2y-3=0 Set them equal to zero

4y=1 2y=3 Add

y=1/4 y=2/3 Divide



This website always helps me when I need help:
http://www.mathonlineteacher.com/

Answer 6

Divide the equation in to two part
(3y-2)^2 - (y+1)^2

Then expand each of them by using (a+b)^2=a^2+2ab+b^2 rule, to avoid mistake dont remove the brackets in this stage.
(9y^2-12y+4) - (y^2+2y+1)

Now change the sign if needed, make all the outer sign to +.
(9y^2-12y+4) + (-y^2-2y-1)

Then simply remove the plus and open the brackets
9y^2-12y+4 - y^2-2y-1

8y^2-14y+3

Try to factorize it yourself, or use ABC technique for quatratic equation, or ABCD technique for cubic equation.

Then you got : (4y-1)(2y-3)

Refer to the source I give you for further problem.

Answer 7

let a = (3y-2)
let b = (y+1)
Therefore,
(3y-2)^2 - (y+1)^2
= a^2 - b^2
Using a common factoring law:
a^2 - b^2 = (a+b)(a-b)
therefore,
(a+b)(a-b)
=(3y-2 + y +1)(3y - 2 - y -1)
=(4y -1)(2y -3)
Which is your desired answer.

Answer 8

this is of the form:
a^2 - b^2 = (a - b)(a + b)
so
(3y - 2 - y -1)(3y - 2 + y + 1) =
(2y - 3)(4y - 1)

Answer 9

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

a=3y-2
b=y+1

[ 3y-2 -(y+1) ] * [ 3y-2 + y+1 ]
(2y-3)(4y-1)

Answer 10

5x + 3y - 2z = 22. resolve for y. Subtract 5x from the two factors of the equation. 3y - 2z = 22 - 5x. upload 2z to the two factors. 3y = 2z - 5x + 22. Divide the two factors via 3. y = (2z - 5x + 22) / 3. answer