Need help translating standard form quadratic equations into factored form?

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May 29th 2008, 08:26 AM
mathdonkey

Need 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 AM
TKHunny

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 AM
TheEmptySet

Quote:

Originally Posted by mathdonkey

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

y = x² - 8x + 15
y = 3x² + 25x + 42

Thanks. (Wink)

I'm guessing you mean in vertex form, that is, $y=a(x-h)^2+k$

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

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

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

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

so we get

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

Will will now rearrange the terms to get
$y=x^2-8x+16 +15-16$

Grouping the first three terms together and factoring gives

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

So we get

$y=(x-4)^2-1$ 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 $(h,k) \to (4,-1)$ 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

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

and try to adapt the method used above.

(Rock)
May 29th 2008, 09:11 AM
mathdonkey

Quote:

Originally Posted by TKHunny

x² - 8x + 15 = (x - 3)(x - 5)

How do I know it's (x - 3)(x - 5) and not (x - 5)(x - 3)?

Quote:

Originally Posted by TKHunny

You do the other one.

In the other one the coefficient of x² is 3. This will change how it's done, so it would be more helpful to see someone else do it and then I can learn from their method.
May 29th 2008, 09:34 AM
Isomorphism

Quote:

Originally Posted by mathdonkey

How do I know it's (x - 3)(x - 5) and not (x - 5)(x - 3)?

Nice joke (Rofl)
May 29th 2008, 09:48 AM
masters

Quote:

Originally Posted by mathdonkey

How do I know it's (x - 3)(x - 5) and not (x - 5)(x - 3)?

In the other one the coefficient of x² is 3. This will change how it's done, so it would be more helpful to see someone else do it and then I can learn from their method.

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

$3x^2+25x+42$

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

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:

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

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

$3x(x+6)+7(x+6) = (3x+7)(x+6)$
May 29th 2008, 09:59 AM
mathdonkey

Quote:

Originally Posted by masters

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

$3x^2+25x+42$

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

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:

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

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

$3x(x+6)+7(x+6) = (3x+7)(x+6)$

Thanks, that helped a lot!

Quote:

Originally Posted by masters

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

Oops, forgot to ask this:

Wont swapping the binomials change the way the parabola looks?
May 29th 2008, 11:04 AM
masters

Swapping the binomials will not have an effect. However, if you were to multiply $ax^2+bx+c=0$ 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 $y=3x^2+25x+42$ in vertex form? You can determine the zeros from the factored version in my previous post. They are $\{-6,\frac{-7}{3}\}$
May 29th 2008, 04:53 PM
mathdonkey

Quote:

Originally Posted by TKHunny

15 > 0, so 3 and 5 need the same sign.
-8 < 0, so both must be negative.

These are good rules to know, so:

What if 15 was a negative number, how would it affect the signs?

What would happen if 8 was a positive number?
May 29th 2008, 05:33 PM
mathdonkey

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 PM
TKHunny

Quote:

Originally Posted by mathdonkey

These are good rules to know, so:

What if 15 was a negative number, how would it affect the signs?

What would happen if 8 was a positive number?

-15 < 0, so factors must have opposite signs.

If they've opposite signs, we're barking up the wrong tree since 5-3 = 2, not 8.
May 29th 2008, 06:03 PM
TKHunny

Quote:

Originally Posted by mathdonkey

y = 20x² - 3x - 9

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 PM
mathdonkey

Sorry TKHunny, I'm a little confused by your method.

Quote:

Originally Posted by TKHunny

Just make a catalogue.

20

1*20
2*10
4*5

9

1*9
3*3

-9 < 0, so opposite signs

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!!!!

And then when I'm finished do I do this to confirm my answer?
May 30th 2008, 04:09 AM
masters

Quote:

Originally Posted by mathdonkey

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?

In your third step you should've factored a 4x instead of 2.

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

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

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 AM
mathdonkey

Quote:

Originally Posted by masters

In your third step you should've factored a 4x instead of 2.

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

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

Once again, thank you! This helped so much.