# Math Help - Factorising

1. ## Factorising

Hello i need some help factorising the following quadratic equation in order to solve it, i have a few to do but i will post an example, then if working out is shown i should be able to do the rest,

2xsquared -5x - 7 = 0

Thanks....

2. Originally Posted by duckegg911
Hello i need some help factorising the following quadratic equation in order to solve it, i have a few to do but i will post an example, then if working out is shown i should be able to do the rest,

2xsquared -5x - 7 = 0

Thanks....
Hello,

I assume that you are familiar with the 3 binomial formulae which are used here:

$2x^2-5x-7 = 0~\iff~2(x^2-\frac52 x - \frac72)=0$

$2(x^2-\frac52 x - \frac72)=0~\iff~ 2\left(x^2-\frac52 x+\frac{25}{16}-\frac{25}{16}-\frac72\right)=0$
$
2\left(\left(x-\frac54\right)^2-\frac{81}{16}\right)=0~\iff~2\left(x-\frac54 + \frac94\right) \cdot \left(x-\frac54-\frac94\right)=0$

Now use the property: A product equals zero if one of the factors equals zero. Thus:

$\left(x+1\right)=0 ~\vee~\left(x-\frac72\right) =0$

And now solve for x.

3. ## Mmmm

There must be an easier way than this...

4. unfortunately there's no
easier way than this
, that is of corse if you don't know the roots of the polynomial beforehand.
Actually I don't know why do you insist on factorizing the polynomial, when a simple formula already exists for findingg the roots of a quadratic equation.

$\begin{array}{l}
p(x) = ax^2 + bx + c \\
x_{1,2} = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
\end{array}$

$p(x) = a(x - x_1 )(x - x_2 )$

Vieta's Formulas -- from Wolfram MathWorld

5. Originally Posted by duckegg911
Hello i need some help factorising the following quadratic equation in order to solve it, i have a few to do but i will post an example, then if working out is shown i should be able to do the rest,

2xsquared -5x - 7 = 0

Thanks....
$2x^2 - 5x - 7 = 0$

Here's another method that works, but only does so if the roots of the equation are rational numbers or integers.

This is called the "ac" method.

Multiply the leading coefficient by the constant coefficient. In this case $2 \cdot -7 = -14$

Now write all the pairs of factors of -14:
1, -14
2, -7
7. -2
14, -1

Now look for a pair that sums to the coefficient of the linear term, in this case -5. Note that 2 + (-7) = -5. (If such a pair does not exist, then the equation cannot be factored in terms of rational numbers.)

So we want to write the linear term as
[tex]-5x = (2 - 7)x = 2x - 7x

Thus
$2x^2 - 5x - 7 = 0$

$2x^2 + 2x - 7x - 7 = 0$

Now factor by grouping:
$(2x^2 + 2x) + (-7x - 7) = 0$

$2x(x + 1) + (-7)(x + 1) = 0$

$(2x - 7)(x + 1) = 0$

Now you can solve this by setting each factor independently equal to 0.

-Dan

6. Originally Posted by topsquark
$2x^2 - 5x - 7 = 0$

Here's another method that works, but only does so if the roots of the equation are rational numbers or integers.

This is called the "ac" method.

Multiply the leading coefficient by the constant coefficient. In this case $2 \cdot -7 = -14$

Now write all the pairs of factors of -14:
1, -14
2, -7
7. -2
14, -1

Now look for a pair that sums to the coefficient of the linear term, in this case -5. Note that 2 + (-7) = -5. (If such a pair does not exist, then the equation cannot be factored in terms of rational numbers.)

So we want to write the linear term as
[tex]-5x = (2 - 7)x = 2x - 7x

Thus
$2x^2 - 5x - 7 = 0$

$2x^2 + 2x - 7x - 7 = 0$

Now factor by grouping:
$(2x^2 + 2x) + (-7x - 7) = 0$

$2x(x + 1) + (-7)(x + 1) = 0$

$(2x - 7)(x + 1) = 0$

Now you can solve this by setting each factor independently equal to 0.

-Dan
is the final answer 2xsquared -12x -7