# Thread: factorising expressions.

1. ## 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.

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

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

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.

$\displaystyle 5t^2 +7t =9$

2. 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.

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

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

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.

$\displaystyle 5t^2 +7t =9$
$\displaystyle cd + 6ce$

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

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

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

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

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

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

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

$\displaystyle 5a + ax - 2b + by$

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...

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Can you tell me what $\displaystyle s$ has to be?

3. 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'?

4. 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 $\displaystyle s$ 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...