# Factors of

#### Casio

##### Member
5(x + 4)(x + 2) = x^2 +2x + 4x + 8

5(x + 4)(x + 2) = x^2 + 6x + 8

5(x + 4)(x + 2) = 5x^2 + 30x + 40

This is the point where I am unsure?

5x^2 + 30x + 40

5 is already a prime, so when I think 10 goes into 30 three times and 10 goes into 40 four times, thus I end up with;

x^2 + 3x + 4 have I completed this correct or should I do it this way;

x^2 + 6x + 5

Thanks

#### DeusAbscondus

##### Active member
i got 5(x+4)(x+2)=5x^2+30x+40
factor out 5 ... =x^2+6x+8
factorize : =(x+4)(x+2)
but i haven't done this stuff for years and am just coming back to it;
offered in spirit of tremulous self-diffidence bordering on readiness-to-chuck-it-in-and-go-back-to-taxi-driving
Godfree (at last, I'm free, nothanks to god, i'm freeeeeee at last)

#### DeusAbscondus

##### Active member
sorry bout that:
can't just spirit that 5 away like that can we:

#### Casio

##### Member
In this last part; 5x^2+30x+40, I am thinking that 5 into 30 goes 6 times and that 5 into 40 goes 8, so I end up with;

x^2 + 6x + 8

Anyone dissagree?

#### DeusAbscondus

##### Active member
In this last part; 5x^2+30x+40, I am thinking that 5 into 30 goes 6 times and that 5 into 40 goes 8, so I end up with;

x^2 + 6x + 8

Anyone dissagree?
Yes, I disagree: like I wrote above: you can't just spirit the 5 away like that, not unless you have an equation which equals 0.
As it is, you have an expression: 5x^2+30x+40
If you factor 5 out of this, you have changed the value of the expression.

What are you trying to do with this? Can I have the context? or a question? or an exercise which you wish to do with this?
Otherwise, as it stands, as I say, it is just an expression, not an equation.

#### Casio

##### Member
It's long winded but what I am trying to do is work out the centre of a circle from a method of completing the square.

I can see there is a problem above because the factors I used cannot be used in the quadratic formula, where a negative cannot have a square root value, so back to the drawing board then?

Basically to continue I will have to by trial and error find the factors of 30 and 40, which there can't be that many?

#### SuperSonic4

##### Well-known member
MHB Math Helper
It's long winded but what I am trying to do is work out the centre of a circle from a method of completing the square.

I can see there is a problem above because the factors I used cannot be used in the quadratic formula, where a negative cannot have a square root value, so back to the drawing board then?

Basically to continue I will have to by trial and error find the factors of 30 and 40, which there can't be that many?
What equation are you starting with? The equation of a circle is $(x-a)^2 + (y-b)^2 = r^2$ where (a,b) is the centre and r the radius so we need a y term.

Can you post the original question in full please?

.
edit: generally if you're completing the square you want it in the form $\left(x+\frac{b}{2a}\right)^2 + \frac{b^2-4ac}{4a^2}$. In this case -\frac{b}{2a} is the x-coordinate of the centre

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#### Casio

##### Member
This is what I have done so far?

(a) The equation
x^2 + y^2 + 6x - 8y + 20 = 0

Represents a circle

Circle centre (-3, 4), and the radius r2 = 5

Find the coordinates of any points in which the circle in part (a) intersects the line y = -2x – 2.

Factors are

5(x + 4)(x + 2) = (x2 + 2x + 4x + 8)
5(x2 + 2x + 4x + 8) = 5x2 + 10x + 20x + 40
= 5x2 + 30x + 40
At this point I wanted to use the quadratic formula after factoring above but doesn't work because I keep getting a negative solution?

#### SuperSonic4

##### Well-known member
MHB Math Helper
This is what I have done so far?

(a) The equation
x^2 + y^2 + 6x - 8y + 20 = 0

Represents a circle

Circle centre (-3, 4), and the radius r2 = 5

Find the coordinates of any points in which the circle in part (a) intersects the line y = -2x – 2.

Factors are

5(x + 4)(x + 2) = (x2 + 2x + 4x + 8)
5(x2 + 2x + 4x + 8) = 5x2 + 10x + 20x + 40
= 5x2 + 30x + 40
At this point I wanted to use the quadratic formula after factoring above but doesn't work because I keep getting a negative solution?
Aha, that makes a lot more sense now. The equation of a circle has both x and y in it (as I posted in post 7). What you want to do is complete the square twice in relation to x and y using the equation given to you.

$(x^2+6x)+(y^2-8y)+20 = 0$ . Complete the square on the left hand side to get it into the form $(x-a)^2+(y-b)^2 = r^2$ which will have centre $(a,b)$ and radius $r$