Question

How do you factor the expression \[{x^4} + 6{x^2} - 7\]?

Answer 1

Hint: Here we use a hit and trial method to find one of the zeroes of the given polynomial. Use the concept of zeroes of a polynomial that they give polynomial value equal to zero when substituted in place of variables. Use the obtained value of zero to make one of the factors of the polynomial. Divide the given polynomial by the factor and calculate remaining factors. Proceed in the same manner until you get all the factors.
* A polynomial of degree \[n\] is a polynomial where the variable has highest power \[n\]. It can be written as \[a{x^n} + b{x^{n - 1}} + ..... + cx + d = 0\]
* If \[x = a\] is one of the zeroes of a polynomial\[a{x^n} + b{x^{n - 1}} + ..... + cx + d = 0\] then we divide the polynomial by \[x - a\]to find other zeros.

Complete step by step solution:
We are given the polynomial \[{x^4} + 6{x^2} - 7\]
Let us write \[f(x) = {x^4} + 6{x^2} - 7\] … (1)
Hit and trial method states that we put in that value in place of x which makes the polynomial as 0 by guessing.
Put \[x = - 1\]
\[ \Rightarrow f( - 1) = {( - 1)^4} + 6{( - 1)^2} - 7\]
\[ \Rightarrow f( - 1) = 1 + 6 - 7\]
\[ \Rightarrow f( - 1) = 0\]
Since the value of x as -1 on substitution gives f(x) equal to 0, then we can say -1 is a factor of the polynomial.
\[(x + 1)\] is a factor of the polynomial. …(2)
Put \[x = 1\]
\[ \Rightarrow f(1) = {(1)^4} + 6{(1)^2} - 7\]
\[ \Rightarrow f(1) = 1 + 6 - 7\]
\[ \Rightarrow f(1) = 0\]
Since the value of x as 1 on substitution gives f(x) equal to 0, then we can say 1 is a factor of the polynomial.
\[(x - 1)\]is a factor of the polynomial. …(3)
From equations (2) and (3) we can say \[(x - 1)\] and \[(x + 1)\] divide \[f(x) = {x^4} + 6{x^2} - 7\]
\[ \Rightarrow (x - 1)(x + 1) = {x^2} - 1\] will also divide \[f(x) = {x^4} + 6{x^2} - 7\]
Using long division method
\[{x^2} - 1)\overline {{x^4} + 6{x^2} + 7} ({x^2} + 7\]
          \[\underline {{x^4} - {x^2}} \]
          \[7{x^2} + 7\]
           \[\underline {7{x^2} + 7} \]
                     \[0\]
Therefore, from long division method we write
\[({x^2} - 1).({x^2} + 7) = {x^4} + 6{x^2} + 7\]
So, \[\left( {{x^2} + 7} \right)\] is another factor of the polynomial … (4)
From equations (1), (2) and (3) we can write:
Factors of the polynomial \[{x^4} + 6{x^2} - 7\] are \[(x + 1);(x - 1)\] and \[({x^2} + 7)\]

\[\therefore \] Factors of the polynomial \[{x^4} + 6{x^2} - 7\] are \[(x + 1);(x - 1)\] and \[({x^2} + 7)\]

Note: Students are likely to make mistakes in the process of long division as they forget to change the sign before the next step. Also, note that a polynomial of degree 4 can have at most 4 zeroes, but it is not necessary that there should be 4 zeros. Keep in mind, here we don’t write the value of factors using complex numbers as we have to write the polynomial using factor form i.e. multiplication of two or more factors, so we can stop after we get two factors.