Can you Check This please?

Need someone to check this and go over it with me please.

*Equation in factored form for a polynomial?*

1. Determine an equation in factored form for a polynomial function with zeros -1 (order 2) and 3 (order 3) that passes through the point (4,5).

y = a [ x + 1 ]² [ x - 3 ]^3 **with 5 = a [ 5²][1^3]**

I don't quite understand the bold part. How Do I incorporate it into my final function/answer.

Am I solving for a?

....5=a(25)(1)

....5=a(25)

....1/5=a

?

2. Determine an equation in factored form for the polynomial function with zeros 2 (order 2), 2/3 and 3 that passes through the point (1,6).

y = a [ x - 2 ]² [ 3x - 2 ][ x - 3 ] ...**with 6 = a [ 1-2]² [ 3 - 2 ] [ 1 - 3 ]**

How do we go from (the green) 2/3 to [3x-2]? And again, i'm confused with the bold part and how to put it into a?

Re: Can you Check This please?

Quote:

Originally Posted by

**tdotodot** Need someone to check this and go over it with me please.

*Equation in factored form for a polynomial?*

1. Determine an equation in factored form for a polynomial function with zeros -1 (order 2) and 3 (order 3) that passes through the point (4,5).

y = a [ x + 1 ]² [ x - 3 ]^3 **with 5 = a [ 5²][1^3]**

I don't quite understand the bold part. How Do I incorporate it into my final function/answer.

Am I solving for a?

....5=a(25)(1)

....5=a(25)

....1/5=a

?

Yes, that's correct. Now just replace "a" in the formula "y = a [ x + 1 ]² [ x - 3 ]^3" with that number.

Do you see how "y = a [ x + 1 ]² [ x - 3 ]^3" **and** "passes through the point (4, 5)" results in "5= a[5^2][1]^3"?

Quote:

2. Determine an equation in factored form for the polynomial function with zeros 2 (order 2), 2/3 and 3 that passes through the point (1,6).

y = a [ x - 2 ]² [ 3x - 2 ][ x - 3 ] ...**with 6 = a [ 1-2]² [ 3 - 2 ] [ 1 - 3 ]**

How do we go from (the green) 2/3 to [3x-2]? And again, i'm confused with the bold part and how to put it into a?

Saying that x= 2/3 is a zero means that it has a factor of the form "x- 2/3". Multiply through by 3.

What do you get if you **do the arithmetic** indicated in "6= a[1- 2]^3 [3- 2][1- 3]"?

Re: Can you Check This please?

Quote:

Originally Posted by

**HallsofIvy** Yes, that's correct. Now just replace "a" in the formula "y = a [ x + 1 ]² [ x - 3 ]^3" with that number.

Do you see how "y = a [ x + 1 ]² [ x - 3 ]^3" **and** "passes through the point (4, 5)" results in "5= a[5^2][1]^3"?

Saying that x= 2/3 is a zero means that it has a factor of the form "x- 2/3". Multiply through by 3.

What do you get if you **do the arithmetic** indicated in "6= a[1- 2]^3 [3- 2][1- 3]"?

#1.

...............So my final answer would then be; y = 1/5 [ x + 1 ]² [ x - 3 ]^3 ?

...............5 = a [5^2][1]^3.... Why are you squaring 5? Why are you putting "[1]^3"? How come the "4" is not incorporated?

#2.

6= a[1- 2]^3 [3- 2][1- 3]

6= a(-1)^3 (1) (-2)

6= a(1)(1)(-2)

6= a(-2)

-3= a

So then... y = a [ x - 2 ]² [ 3x - 2 ][ x - 3 ]

.............. y= -3[ x - 2 ]² [ 3x - 2 ][ x - 3 ] ...would be my final answer?

Re: Can you Check This please?

Quote:

Originally Posted by

**tdotodot** #1.

...............So my final answer would then be; y = 1/5 [ x + 1 ]² [ x - 3 ]^3 ?

...............5 = a [5^2][1]^3.... Why are you squaring 5? Why are you putting "[1]^3"? How come the "4" is not incorporated?

The "4" **is** incorporated- that is where the "5" and "3" came from. You had

y= a[x+ 1]^2[x- 3]^3 and you want y= 5 when x= 4 . Replacing y by 5 and x by 4 in the equation

5= a[4+ 1]^2[4- 3]^3= a[5]^2[1]^3.

Quote:

#2.

6= a[1- 2]^3 [3- 2][1- 3]

This should be [1- 2]^2, not to third power. Other than that, yes setting x= 1, y= 6 is correct.

Quote:

6= a(-1)^3 (1) (-2)

6= a(1)(1)(-2)

(-1)^3= -1, not 1 but since that was supposed to be squared, this is correct!

Quote:

6= a(-2)

-3= a

So then... y = a [ x - 2 ]² [ 3x - 2 ][ x - 3 ]

.............. y= -3[ x - 2 ]² [ 3x - 2 ][ x - 3 ] ...would be my final answer?

Yes. You **could**, rather than have represented the "zero at 2/3" by 3x- 2 have used "x- 2/3" instead.

That would give you y= a[x- 2]^2[x- 2/3][x- 3] and setting x= 1, y= 6

6= a(1- 2)^2(1- 2/3)(1- 3)= a(1)(1/3)(-2)= -(2/3)a

to -(2/3)a= 6 and now a= -6(3/2)= -9.

That gives y= -9(x- 2)^2(x- 2/3)(x- 3) but factoring 9 into 3(3) and multiplying one of those "3"s into the (x- 2/3) term,

y= -3(x- 2)^2(3x- 2)(x- 3) exactly the same as before!