Factor the polynimial....please help!!!?

Question

I really would love your help I'm so confused on how to do this!!

4x^3-28x^2+49x



12y^3-36y^2+27y

Answer 1

1) 4x³-28x²+49x

factor the x

x(4x²-28x+49)

x(2x-7)²

2) 12y³-36y²+27y

factor 3y

3y(4y²-12y+9)

3y(2y-3)²

Answer 2

4x^3 - 28x^2 + 49

Factor out an x.

x(4x^2 - 28x + 49)

Using the ac technique for factoring, note that 49 times 4 = 196, and -14 times -14 = 196, while -14x + -14x = -28x, the middle term. Split -28x into -14x and -14x.

x(4x^2 - 14x - 14x + 49)
x( 2x(2x - 7) - 7(2x - 7) )

Group.

x(2x - 7)(2x - 7)

2) 12y^3 - 36y^2 + 27y

Factor out a 3y

3y(4y^2 - 12y + 9)

Using the same "AC" technique, multiply 4 and 9 together, which gives 36. What two factors of 36 give -12? The answer is -6 and -6.

3y ( 4y^2 - 6y - 6y + 9 )
3y ( 2y(2y - 3) - 3(2y - 3) )

3y (2y - 3)(2y - 3)

Answer 3

First of all, we can extract x (common factor)

x(4x^2-28x+49)

Now if you notice that 4 and 49 are perfect squares, try:

x(2x-7)^2

and you'll see that it works.

2)

3y(4y^2-12y+9)

3y(2y-3)^2

Answer 4

4x^3-28x^2+49x
taking x common.....
x(4x^2-28x+49)
splitting the middle term, we get
x(2x(2x-7)-7(2x-7))
so we get x (2x-7)^2



12y^3-36y^2+27y
similarly in dis question also we take 3y common
so we get
3y(4y^2-12y+9)
again splitting the middle term,
3y(4y^2-6y-6y+9)
3y(2y(y-3)-2(y-3))
3y(2y-2)(y-3)

Answer 5

4x^3-28x^2+49x
=x(2x-7)^2

12y^3-36y^2+27y
=3y(2y-3)^2

Answer 6

x(4x^2-28x+49)=x(2x-7)(2x-7)=x(2x-7)^2