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cookster · Sep 19, 2009, 01:56 PM

hi could anybody show me how to factorise this quadratic please.
3x^+5x^+2=0
thanks for any help.

Unknown008 · Sep 19, 2009, 09:20 PM

I guess you mean $3x^2 + 5x - 2=0$ ?

I'll go through it the method I use.

1. First, look for the factors of the coefficient of x^2 and the constant.
$3x^2 = 3x \times x$ or $-3x \times -x$
$-2 = -2 \times 1$ or $2 \times -1$

2. I'll take the first possibilities for 3x^2 and -2.
$3x\,\,\,\,\,x \\ -2\,\,\,\,\,1$

3a. Multiply one term from the upper ones by one of the lower ones, then add both results:
$3x\,\,\,\,\,\,\,\,\,x \\ -2\,\,\,\,\,\,\,\,\,1 \\ \\ -6x\,\,\,\,\,x= -5x$

3b. As you can see, the final result is -5x, which is not the 5x that is present in the original equation. Ok, multiply the terms diagonally:
$3x\,\,\,\,\,\,\,\,\,x \\ -2\,\,\,\,\,\,\,\,\,1 \\ \\ -2x\,\,\,\,\,3x= x$

4a. As you can see, the final result is -5x, which is not the 5x that is present in the original equation. Ok, change the factors (from the ones in 1.) . Use the first and second ones.
$3x\,\,\,\,\,x \\ 2\,\,\,\,\,-1$

4b. Do the same thing as in step 3a.
$3x\,\,\,\,\,\,\,\,\,x \\ 2\,\,\,\,\,\,\,\,\,-1 \\ \\ 6x\,\,\,\,\,-x= 5x$

Here! You got the 5x that you had in the original equation. So, Bracket the diagonal terms together.
$(3x-1)(x+2)$

If you expand, you'll have the original expression back.

Now, rewrite everything:
$3x^2 + 5x - 2 = (3x-1)(x+2) = 0$

Since any number multiplied by zero gives zero, either (3x-1) is zero, or (x+2) is zero. You have two equations then:
$3x-1 = 0$ or $x+2 =0$

Solve for both. Tell us what you get.

I hope it helped! :)