Could someone please translate these equations and explain the steps for doing it?

y = x² - 8x + 15

y = 3x² + 25x + 42

Thanks. (Wink)

- May 29th 2008, 08:26 AMmathdonkeyNeed help translating standard form quadratic equations into factored form?
Could someone please translate these equations and explain the steps for doing it?

y = x² - 8x + 15

y = 3x² + 25x + 42

Thanks. (Wink) - May 29th 2008, 08:59 AMTKHunny
It's not magic. You just have to look.

x² - 8x + 15 =

Obvious part:

(x __ ___)(x __ ___)

Factors of 15

1,15

3,5

That's it.

Can one get 8 from these factors?

1+15 = 16 -- Nope.

1-14 = -14 -- Nope

3-5 = -2 -- Nope

3+5 = 8 -- Yes!!

(x __ 3)(x __ 5)

15 > 0, so 3 and 5 need the same sign.

-8 < 0, so both must be negative.

x² - 8x + 15 = (x - 3)(x - 5)

You do the other one. - May 29th 2008, 09:01 AMTheEmptySet
I'm guessing you mean in vertex form, that is, $\displaystyle y=a(x-h)^2+k$

To do the we need to complete the square, to do this we take half of the linear coeffeint and square it.

$\displaystyle y=x^2-8x+15$

So the linear coeffeint is $\displaystyle -8$ half of it is $\displaystyle \frac{-8}{2}=-4$ and now we square it to get $\displaystyle (-4)^2=16$ This is what we need to complete the square.

We will now add zero to the euqation in the form of $\displaystyle 0=(16-16)$

so we get

$\displaystyle y=x^2-8x+15+0=x^2-8x+15+(16-16)$

Will will now rearrange the terms to get

$\displaystyle y=x^2-8x+16 +15-16$

Grouping the first three terms together and factoring gives

$\displaystyle y=[x^2-8x+16]-1=(x-4)^2-1$

So we get

$\displaystyle y=(x-4)^2-1$ We can now identify a, h, and k

a=1 h=4 and k=-1

So we know that the vertext is located at the point $\displaystyle (h,k) \to (4,-1)$ and the the parabola opens up because a is postitive.

I hope this helps and is what you wanted.

If so here is a hint for #2 factor like this

$\displaystyle y=ax^2+bx+c=a(x^2+\frac{b}{a}x)+c$

and try to adapt the method used above.

(Rock) - May 29th 2008, 09:11 AMmathdonkey
How do I know it's (x - 3)(x - 5) and not (x - 5)(x - 3)?

Quote:

Originally Posted by**TKHunny**

- May 29th 2008, 09:34 AMIsomorphism
- May 29th 2008, 09:48 AMmasters

Multiplication is commutative, so (x - 3)(x - 5) = (x - 5)(x - 3)

$\displaystyle 3x^2+25x+42$

To factor this trinomial, first multiply leading coef. times constant: $\displaystyle 3\times42$

Result: 126

Find two factors of 126 that add up to the middle term's coef. of 25.

That'd be 18 and 7.

Replace the middle term with these two factors:

$\displaystyle 3x^2+18x+7x+42$

Factor by grouping: $\displaystyle (3x^2+18x)+(7x+42)$

$\displaystyle 3x(x+6)+7(x+6) = (3x+7)(x+6)$ - May 29th 2008, 09:59 AMmathdonkey
- May 29th 2008, 11:04 AMmasters
Swapping the binomials will not have an effect. However, if you were to multiply $\displaystyle ax^2+bx+c=0$ by -1, then the parabola would open in the opposite direction. The vertex would change, but the x-intercepts would be the same.

Did you need to see $\displaystyle y=3x^2+25x+42$ in vertex form? You can determine the zeros from the factored version in my previous post. They are $\displaystyle \{-6,\frac{-7}{3}\}$ - May 29th 2008, 04:53 PMmathdonkey
- May 29th 2008, 05:33 PMmathdonkey
I'm having some trouble with this question too.

y = 20x² - 3x - 9

I get this:

= 20x² + 12x - 15x - 9

= (20x² + 12x) (-15x - 9)

= 2(10x + 12x) 3(-5x - 3)

(I don't know what to do next)

But the answer is supposed to be:

y = (4x - 3) (5x + 3)

What am I doing right/wrong? - May 29th 2008, 05:57 PMTKHunny
- May 29th 2008, 06:03 PMTKHunny
Just make a catalogue.

20

1*20

2*10

4*5

9

1*9

3*3

-9 < 0, so opposite signs

We're looking for a -3

1*1 - 9*20 -- too big

1*20 - 9*1 = 11

1*3 - 9*3 = -26

2*1 - 10*20 -- too big

2*20 - 10*1 = 30

2*3 - 10*3 = -27

4*1 - 5*20 -- too big

4*20 - 5*1 = 75

4*3 - 5*3 = -3 -- Aha!!!!

(4x - 3)(5x + 3)

Be systematic.

Note: I started witht the most extreme values, just for the practice. I don't recommend that if you REALLY want the quickest path to the answer. - May 29th 2008, 07:02 PMmathdonkey
Sorry TKHunny, I'm a little confused by your method.

Using those numbers I can write (4x - 3) (5x + 3). Is that all I have to do?

Quote:

We're looking for a -3

1*1 - 9*20 -- too big

1*20 - 9*1 = 11

1*3 - 9*3 = -26

2*1 - 10*20 -- too big

2*20 - 10*1 = 30

2*3 - 10*3 = -27

4*1 - 5*20 -- too big

4*20 - 5*1 = 75

4*3 - 5*3 = -3 -- Aha!!!!

- May 30th 2008, 04:09 AMmasters
In your third step you should've factored a 4x instead of 2.

$\displaystyle y=4x(5x+3)-3(5x+3)$

$\displaystyle y=(4x-3)(5x+3)$

As far as those other questions, it is best if you get a graphing calculator and try those different transformations you're talking about. - May 30th 2008, 06:10 AMmathdonkey