Factorising Quadratics

**Ziandia** · February 11th 2008, 08:10 PM

Hi all, I started this subject today and I have many questions of a similar nature. It would be much appreciated if someone could give me a step-by-step tutorial on how to answer a question such as this:

Factorize these quadratic expressions:
3x^2 - 7x + 2

or

9x^2 + 30x + 25

I've never used this forum before so I'm sorry if I did anything wrong.
Thanks heaps!
Sam

**topher0805** · February 11th 2008, 08:14 PM

The easiest way is to set the expressions equal to 0 then find the roots using the quadratic equation.

Do you know the quadratic equation?

**Ziandia** · February 11th 2008, 08:26 PM

I don't sorry

**topher0805** · February 11th 2008, 08:48 PM

Ok, your equations are:

$3x^2 - 7x + 2$

and:

$9x^2 + 30x + 25$

They are of the form:

$ax^2 + bx + c$ where $a$ , $b$ , and $c$ are all constants.

The quadratic equation is:

$x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$

So, you plug in the numbers, and voila! You have the roots of the expression.

**angel.white** · February 11th 2008, 09:01 PM

Originally Posted by Ziandia

Hi all, I started this subject today and I have many questions of a similar nature. It would be much appreciated if someone could give me a step-by-step tutorial on how to answer a question such as this:

Factorize these quadratic expressions:
3x^2 - 7x + 2

or

9x^2 + 30x + 25

I've never used this forum before so I'm sorry if I did anything wrong.
Thanks heaps!
Sam

I learned this 2 semesters ago, so forgive me if I err (someone feel free to correct any misconceptions I have, I tried to look it up but didn't find anything)

Look at the polynomial 3x^2 - 7x + 2

The first coefficient is 3, and the last is 2. There is a property that if you can remove a factor, it will take the form of (x+p/q) or (x+q/p) where p is a factor of the first coefficient, and q is a factor of the second coefficient. So potential factors are $(x+\frac 12)$ , $(x+\frac 11)$ , $(x+\frac {-1}2)$ , $(x+\frac {-1}2)$ , $(x+\frac 32)$ , $(x+\frac 31)$ , $(x+\frac {-3}2)$ , $(x+\frac {-3}1)$ , $(x+\frac 13)$ , $(x+\frac {-1}3)$ , $(x+\frac 23)$ , $(x+\frac {-2}3)$ , $(x+\frac 21)$ , $(x+\frac {-2}1)$

In this case, I found that $(x+\frac {-2}1) = (x-2)$ works.

To pull it out, I used division:

Perform the division:

$\begin{array}{rr}<br /> &3x-1\\<br /> (x-2) & \overline{\begin{array}{|r}3x^2-7x+2\end{array}}<br /> \end{array}$

So 3x^2 - 7x + 2 can be factored into (x-2)(3x-1)

----------

The second one is pretty easy, because it's obvious. The first and last coefficients are perfect squares, so you could look at it like this:
$9x^2 + 30x + 25 = (3x)^2+30x+5^2$

From here it should be rather clear you are looking at a perfect square, if it is not, try playing around with perfect squares a bit so that you can learn to recognize them.
$(3x)^2+30x+5^2 = (3x+5)^2$

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