Factorising quadratics with leading coefficient not 1IB MYP Maths Extended: Revision notes

To factorise $ax^2+bx+c$ with $a\neq1$, first take out any common factor. Then find two numbers that multiply to $ac$ and add to $b$, split the middle term, and factorise in pairs. Example: $6x^2+7x-3$. Here $ac=-18$ and $b=7$. The numbers are $9$ and $-2$. So $6x^2+9x-2x-3=3x(2x+3)-1(2x+3)=(3x-1)(2x+3)$. Always expand to check: the outer and inner products must add to $b$.

A difference of two squares has no middle term: $p^2x^2-q^2=(px-q)(px+q)$. For example $9x^2-16=(3x-4)(3x+4)$. If no pair of whole numbers multiplies to $ac$ and adds to $b$, the quadratic does not factorise neatly. Use the quadratic formula instead.

Rearrange to the form $ax^2+bx+c=0$, factorise, then use the zero product rule: if $(px+q)(rx+s)=0$ then $px+q=0$ or $rx+s=0$. Example: $2x^2+5x-75=0$ gives $(2x+15)(x-5)=0$, so $x=-7.5$ or $x=5$. A quadratic can have two solutions, so give both unless the context rejects one.

For $ax^2+bx+c=0$: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$ The discriminant $b^2-4ac$ tells you how many solutions: positive gives two, zero gives one repeated solution, negative gives none. If it is a perfect square, the quadratic factorises. Example: $5t^2-4t-2=0$ gives $t=\frac{4\pm\sqrt{56}}{10}=1.15$ or $-0.348$.

In real problems, check each solution makes sense. A negative time or length must be rejected. Write a sentence that answers the question, with units. For a downward parabola such as $P=-2x^2+11x-12$, the roots are where $P=0$, the maximum is halfway between them, and $P>0$ between them.