# factorising expressions.

• Dec 31st 2009, 06:09 AM
tubruk
factorising expressions.
Hi, I'm new to the forums and pretty much fail at maths.
General (Basic) maths I can do, but algebra etc. is alien to me.
I'm looking for some help in factorising the following expressions.

$cd + 6ce$
$5x^4 - 3x^3 + 4x^2$
$5a + ax -2b + by$

and finding the solution to this equation through factorisation:
$4s^2 -11s -3 =0$

I also have another problem regarding solving an equation using the Formula Method?
I'll include the question in case it will help you, help me identify the formula.

$5t^2 +7t =9$
• Dec 31st 2009, 06:25 AM
Prove It
Quote:

Originally Posted by tubruk
Hi, I'm new to the forums and pretty much fail at maths.
General (Basic) maths I can do, but algebra etc. is alien to me.
I'm looking for some help in factorising the following expressions.

$cd + 6ce$
$5x^4 - 3x^3 + 4x^2$
$5a + ax -2b + by$

and finding the solution to this equation through factorisation:
$4s^2 -11s -3 =0$

I also have another problem regarding solving an equation using the Formula Method?
I'll include the question in case it will help you, help me identify the formula.

$5t^2 +7t =9$

$cd + 6ce$

Can you see that each term has been multiplied by $c$?

That means that you can take $c$ out as a common factor.

So $cd + 6ce = c(d + 6e)$.

$5x^4 - 3x^3 + 4x^2$

Can you see that each term has $x^2$ in it? (note that $x^4 = x^2\cdot x^2$ and $x^3 = x\cdot x^2$)

So we can take $x^2$ out as a common factor.

$5x^4 - 3x^3 + 4x^2 = x^2(5x^2 - 3x + 4)$.

$5a + ax - 2b + by$

Can you see anything that goes into every term in this expression? I can't...

The best you could hope for in this case is to factorise groups of two...

$5a + ax - 2b + by = a(5 + x) + b(-2 + y)$.

We still don't get any common factors though...

$4s^2 -11s -3 =0$

This is a quadratic equation of the form $ax^2 + bx + c = 0$.

We can factorise the quadratic by finding two numbers that multiply to be $ac$ and add to be $b$.

$ac = -12$ and $b = -11$.

What numbers multiply to be $-12$ and add to be $-11$? Surely the two numbers must be $-12$ and $1$.

Watch what happens when I break up the middle term...

$4s^2 - 11s - 3 = 0$

$4s^2 - 12s + 1s - 3 = 0$.

If we factorise by groups of two, we can find a common factor...

$4s(s - 3) + 1(s - 3) = 0$

Can you see that $s - 3$ is a common factor?

$(s - 3)(4s + 1) = 0$.

Now we have two "things" multiplying to become $0$. This means at least one of those "things" had to be $0$.

So $s - 3 = 0$ or $4s + 1 = 0$.

Can you tell me what $s$ has to be?
• Dec 31st 2009, 06:47 AM
tubruk
Thank you for your Quick Reply and thorough Help.

Yes, 's' has to be 3 in the case of 's -3 =0'
But can't it also be '-0.25' if its '4s+1=0'?
• Dec 31st 2009, 06:53 AM
Prove It
Quote:

Originally Posted by tubruk
Thank you for your Quick Reply and thorough Help.

Yes, 's' has to be 3 in the case of 's -3 =0'
But can't it also be '-0.25' if its '4s+1=0'?

Yes that's exactly right. There are two values that $s$ could be to make the equation true.

Check it by putting those values into the original equation and see if you get 0 each time...