Question

How do you factor ${{t}^{4}}-{{t}^{2}}$completely ?

Answer 1

Hint: To factorize the given equation first we can take ${{t}^{2}}$ common from the equation, we know that ${{t}^{2}}$ is a factor of ${{t}^{4}}$ and ${{t}^{2}}$ is a factor of itself. So we can take ${{t}^{2}}$ common from the equation. The we can apply the formula ${{a}^{2}}-{{b}^{2}}$ to solve the equation. We know that ${{a}^{2}}-{{b}^{2}}$ is equal to (a + b) (a – b) .

Complete step by step solution:
The given equation is ${{t}^{4}}-{{t}^{2}}$
We can take ${{t}^{2}}$ common form the above equation , so by taking ${{t}^{2}}$ common we get
$\Rightarrow {{t}^{4}}-{{t}^{2}}={{t}^{2}}\left( {{t}^{2}}-1 \right)$
We know the algebraic formula ${{a}^{2}}-{{b}^{2}}$ is equal to (a + b) (a – b) .
We can apply this formula to solve ${{t}^{2}}-1$ , we can write ${{t}^{2}}-1$ as (t + 1) (t – 1) , replacing it in the equation we get
$\Rightarrow {{t}^{4}}-{{t}^{2}}={{t}^{2}}\left( t+1 \right)\left( t-1 \right)$
So we can see that ${{t}^{2}}\left( t+1 \right)\left( t-1 \right)$ is the factor form of the equation ${{t}^{4}}-{{t}^{2}}$

Note: We can find the roots of any polynomial equation from its factor form. If the factor form of any polynomial function f (x) is (x – a) (x – b) (x – c) (x – d) then a, b, c, and d are roots of the equation f (x) . We have evaluate the factor form of ${{t}^{4}}-{{t}^{2}}$ is ${{t}^{2}}\left( t+1 \right)\left( t-1 \right)$ , so the roots of the equations ${{t}^{4}}-{{t}^{2}}$ are 0, -1 and 1 where we can see 0 is the repeated root. All the roots of ${{t}^{4}}-{{t}^{2}}$ are real. If any constant c satisfies the polynomial f (x) = 0 the (x – c) is a factor of f (x).