Question

Factorise the given expression ${x^4} + {x^2} - 6$?

Answer 1

Hint: The given problem requires us to factorise a polynomial. The given polynomial is of degree $4$ but can be reduced to a simple quadratic polynomial using a substitution. There are various methods that can be employed to factorise a quadratic polynomial like completing the square method, using quadratic formula and by splitting the middle term.

Complete step by step answer:
Consider the given polynomial $p(x) = {x^4} + {x^2} - 6$
The polynomial is of degree $4$. We can convert the polynomial into a quadratic polynomial by a simple substitution.
Let ${x^2} = z$, then, ${x^4} = {\left( {{x^2}} \right)^2} = {z^2}$
Now the variable is changed from x to z. Now, the given polynomial has become $p(z) = {z^2} + z - 6$
Now, quadratic polynomials can be factored using various methods like completing the square method, using quadratic formulas and by splitting the middle term. We would factorise the polynomial using splitting the middle term.

Hence, the middle term has to be split into two terms such that the sum of the terms gives us the original middle term and product of the terms gives us the product of the constant term and coefficient of ${x^2}$.
So, $p(z) = {z^2} + z - 6$
\[ = {z^2} + (3 - 2)z - 6\]
Opening the brackets and simplifying further, we get,
$ = {z^2} + 3z - 2z - 6$
$ = ({z^2} + 3z) - (2z + 6)$
Taking out the common terms from both the brackets,
$ = z(z + 3) - 2(z + 3)$
$ = \left( {z + 3} \right)\left( {z - 2} \right)$
Now, changing back the variable from z to x, as given variable was x. We get,
$p(x) = ({x^2} + 3)({x^2} - 2)$

Note: Similar to quadratic polynomials, quadratic solutions can also be solved using factorisation method. Besides factorisation, there are various methods to solve quadratic equations such as completing the square method and using the Quadratic formula. Splitting of the middle term can be a tedious process at times when the product of the constant term and coefficient of ${x^2}$ is a large number with a large number of divisors. Special care should be taken in such cases.