Question

How do you factor the polynomial completely ${x^4} - 24{x^2} - 25$?

Answer 1

Hint: Here we need to factorize the given polynomial ${x^4} - 24{x^2} - 25$ . Note that the degree of the polynomial is 4. We convert the polynomial into quadratic polynomial by taking $t = {x^2}$, so now it is a quadratic polynomial. Firstly, we find the pair of integers whose product is -25 and whose sum is -24. Then for the obtained expression, we substitute back $t = {x^2}$. Now we note that one of the pairs is a perfect square. We simplify it by making use of the difference of square formulas. Then we solve it to obtain the required result.

Complete step-by-step solution:
Given the polynomial of the form,
${x^4} - 24{x^2} - 25$ …… (1)
We are asked to factorize the above polynomial completely which is given by the equation (1).
Note that the degree of the above polynomial is 4.
Firstly, we try to write the above polynomial as a quadratic polynomial, so that it is easier to simplify.
We rewrite ${x^4}$as ${x^4} = {\left( {{x^2}} \right)^2}$
Thus, the equation (1 becomes,
$ \Rightarrow {\left( {{x^2}} \right)^2} - 24{x^2} - 25$ …… (2)
Now we take ${x^2}$ as some variables say t. Let $t = {x^2}$. Substituting this in the equation (2) we get,
$ \Rightarrow {t^2} - 24t - 25$
This equation is of the form $a{x^2} + bx + c$. We solve this type of equation by finding a pair of integers whose product is $a \cdot c$ and whose sum is $b$.
Here $a = 1,$ $b = - 24$ and $c = - 25$
In this problem, the product is,
 $a \cdot c = 1 \cdot ( - 25)$
$ \Rightarrow a \cdot c = - 25$
The sum is, $b = - 24$
$ \Rightarrow b = - 25 + 1$
Now we write the factored form using these integers, we get,
$ \Rightarrow {t^2} + t - 25t - 25$
Factor out the common terms, we get,
$ \Rightarrow t(t + 1) - 25(t + 1)$
Now factoring out the common factors we get,
$ \Rightarrow (t - 25)(t + 1)$
Now we replace $t = {x^2}$, we get,
$ \Rightarrow ({x^2} - 25)({x^2} + 1)$
This can also be written as,
$ \Rightarrow ({x^2} - {5^2})({x^2} + 1)$ …… (3)
Note that in the first pair both terms are perfect squares and is of the form ${a^2} - {b^2}$.
We factor it using difference of square formula given by,
${a^2} - {b^2} = (a + b)(a - b)$
Note that here $a = x$ and $b = 5$.
Hence we get,
$ \Rightarrow {x^2} - {5^2} = (x + 5)(x - 5)$
Substituting this in the equation (3), we get,
$ \Rightarrow (x + 5)(x - 5)({x^2} + 1)$
Hence the factorization of the equation ${x^4} - 24{x^2} - 25$ is given by $(x + 5)(x - 5)({x^2} + 1)$.

Note: If the degree of the polynomial is 4, firstly convert the polynomial into quadratic polynomial, so that it makes our simplification easier. Students must know the perfect squares of the numbers. Otherwise it becomes difficult to solve this kind of problem.
They must know the following formulas,
(1) ${(a + b)^2} = {a^2} + 2ab + {b^2}$
(2) ${(a - b)^2} = {a^2} - 2ab + {b^2}$
(3) ${a^2} - {b^2} = (a + b)(a - b)$