# Can you Check This please?

• Sep 15th 2013, 08:44 AM
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
?

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?
• Sep 15th 2013, 01:27 PM
HallsofIvy
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]"?
• Sep 15th 2013, 04:14 PM
tdotodot
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?
• Sep 15th 2013, 05:41 PM
HallsofIvy
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!