Factorisation of a not straight-forward expression

Question

I am completely lost with this question:

Factorise 3x ²- 11x + 6

How do you factorise this?

Please help

Answer 1

Hint : $3x^2-11x+6=(3x^2-9x)-(2x-6)$

But this type of problem is usually solved with the cross method.

Answer 2

Compute the Solutions of $$x^2-\frac{11}{3}x+2=0$$ so $$x_{1,2}=\frac{11}{6}\pm\sqrt{\frac{49}{36}}$$ this gives $$x_{1,2}=\frac{11}{6}\pm \frac{7}{6}$$ and we get $$x_1=3$$ or $$x_2=\frac{2}{3}$$ and we can write $$(x-3)\left(x-\frac{2}{3}\right)=0$$

Answer 3

You can do trial and error, or use what I sort of remember as the $pqrs$ method...

For trial and error: $3x^2$ factors as $3x\cdot x$, so, if it factors, it's of the form $$(3x\qquad)(x\qquad)$$ ;and $6$ factors variously as $\pm6\cdot\pm1$ and $\pm3\cdot \pm2$. You need the middle term to work out. For instance, $6,1$ doesn't work: $$(3x+6)(x+1)$$ has the wrong middle term, $9x$... There's a total of $8$ combinations to try, if you remember that order will matter.

A little experimentation leads to $\mathbf{(3x-2)(x-3)}$.

The second method goes as follows: $(px+r)(qx+s)=pqx^2+(ps+qr)x+rs$. Inspecting this reveals that the $x$ term splits into $2$ terms whose coefficients have product $pqrs$, which is the product of the $x^2$ coefficient, $pq$ and the constant term, $rs$. Thus, if we start with $ax^2+bx+c$, we can look for two numbers whose sum is $b$ and whose product is $ac$.

So, in our example, we need two numbers (integers) whose sum is $-11$, and whose product is $3\cdot6=18$. We get $-2$ and $-9$.

Having found the required two numbers, we use them to split the $x$ term and proceed to factor.

$$\displaystyle 3x^2-11x+6=3x^2-2x-9x+6=x(3x-2)-3(3x-2)=(x-3)(3x-2)$$.

Note, our scratchwork above guarantees that we will always get a common binomial factor (in this case $(3x-2)$).

Final note: either method will work if the quadratic polynomial is in fact factorizable over $\mathbb Z$.