Hello there, I've got a quadratic equation. I'm supposed to solve it by factorising but it's already factorized.

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

Can you help me?

Use the fact that $a\cdot b=0 \Leftrightarrow a=0 \ \mbox{or} \ b=0$

For $(3x + 1)(x + 3) = 19 - 2(x +2)$

Do I expand everything on the left hand side then do the same on the the Right hand side then move the right hand side to the left hand side to make it = 0?

Yep have a go at that.

After a long complex situation I get

$x^2+10x+18=0$

Which I can't do as nothing multiplys to make 18 thats add to 10

For $\displaystyle ax^2+bx+c=0\implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

But I've got to do via factorisation. Don't think I can factorise the above. Think I done something wrong.

Complete the square then.

Originally Posted by JibblyJabbly
But I've got to do via factorisation
$x^2 + 10x + 18 = 0$

if you insist ...

$(x + 5 + \sqrt{7})(x + 5 - \sqrt{7}) = 0$

Originally Posted by skeeter
$x^2 + 10x + 18 = 0$

if you insist ...

$(x + 5 + \sqrt{7})(x + 5 - \sqrt{7}) = 0$
Just to double check
$(3x + 1)(x + 3) = 19 - 2(x +2)$ is equal to

$x^2 + 10x + 18 = 0$ right? I assume I had problems with the Left hand side of the equation

Better check my workings but,

$(3x + 1)(x + 3) = 19 - 2(x +2)$

$3x^2+9x+x+3 = 19 - 2x -4$

$3x^2+10x+3 = 15 - 2x$

$3x^2+12x-12 = 0$

$3(x^2+4x-4) = 0$

$x^2+4x-4 = 0$