Previously on the blog, we wrote an article titled, a quick way to factorise cubics. Now we give you another method, which is also convenient.
Remember, cubics have the form:
$${\normalsize f(x)=ax^3+bx^2+cx+d}$$Using the factor theorem, you will first need to find, \(\normalsize f(x) = 0\). Then multiply it by some quadratic expression \(\,\normalsize (mx^2 +nx+p)\,\). The expression
Let us try a few examples:
-
Factorise \(\normalsize f(x)=x^3-5x^2-8x+12\)
The steps:
- Find the factor or the root, using factor theorem. This means try different values of \(\normalsize x\) that will make \(\normalsize f (\,x)\, =0\). In this case, the root, \(\normalsize x = 1\) will make \(\normalsize f (\,x)\, =0\). The factor is therefore, \((\,\normalsize x-1)\,\)
- Now we must multiply this factor by the quadratic.
$${\normalsize f(\,x)\,=(\,x-1)\,(\,mx^2 +nx+p)\,}$$
The coefficient of the \(\normalsize x^2\) of the quadratic equals the coefficient of \(\normalsize x^3\). Therefore, \(\normalsize m=1\).
Find the sum of \(\normalsize x\) multiplied by \(\normalsize nx\) and \(\normalsize -1\) multiplied by \(\normalsize mx^2\) and equate them to \(bx^2\) of the quadratic.
$$ \normalsize \begin{align*} nx^2 + (-mx^2) &= bx^2 \\ nx^2 -mx^2 &= -5x^2 \\ nx^2 -x^2 &= -5x^2 \\ nx^2 &= -4x^2 \\ n &= -4 \end{align*} $$Multiply \(-1\) by \(p\) and equate it to \(d\).
$$ \normalsize \begin{align*} -p&=d\\ -p&=12\\ p&=-12 \end{align*} $$
Now substitute the values in the expression
$$ \normalsize \begin{align*} f(\,x)\,&= x^3-5x^2-8x+12\\ f(\,x)\, &= (\,x-1)\,(\,x^2-4x-12)\, \\ f(\,x)\, &= (\,x-1)\,(\,x-6)\,(\,x+2)\, \end{align*} $$
The expression has been factorised.
All these can be done in one step. But it gets better only with more practice.
-
Factorise \( \normalsize f(x)=-x^3+10x^2-17x-28\)
The factor is \(\normalsize(\,x+1)\,\).
By following the steps in (1), we get:
$$ \normalsize \begin{gather} m=-1\\ n=11\\ p=-28 \end{gather} $$The expression becomes:
$$ \normalsize \begin{align*} f(\,x)\,&= -x^3+10x^2-17x-28\\ f(\,x)\, &= (\,x+1)\,(\,-x^2+11x-28)\, \\ f(\,x)\, &= (\,x+1)\,[\,-(\,x^2-11x+28)\,]\, \\ f(\,x)\, &= (\,x+1)\,[\,-(\,x-7)\,(\,x-4)\,]\, \\ f(\,x)\, &= -(\,x+1)\,(\,x-6)\,(\,x-4)\, \end{align*} $$
If this method does not work for you, try another one here.