Question

Factorise the given expression: $8 - 4a - 2{a^3} + {a^4}$.

Answer 1

Hint: Observe the given expression carefully. Find out the common factors if any and take it out. Then we get the factors of the given expression. Here we can take $4$ common from the first two terms and ${a^3}$ common from the last two terms. Then we again get a common factor. Taking that too common outside we can factorise the given expression to two factors.

Complete step-by-step answer:
We have to factorise $8 - 4a - 2{a^3} + {a^4}$.
Consider the first two terms and last two terms separately.
We can take $4$ common from the first two terms and ${a^3}$ common from the last two terms.
This gives,
$8 - 4a - 2{a^3} + {a^4} = 4(2 - a) - {a^3}(2 - a)$
Now we can see the factor $2 - a$ is common in both the terms.
We can take it common outside.
So we get,
$8 - 4a - 2{a^3} + {a^4} = (2 - a)(4 - {a^3})$
Thus we get the factors of the expression as $2 - a$ and $4 - {a^3}$.

Additional information:
If while solving we get expressions like ${a^2} - {b^2}$ we can factorise it.
For we have,
${a^2} - {b^2} = (a + b)(a - b)$
Also we have the results,
${(a + b)^2} = {a^2} + 2ab + {b^2}$
${(a - b)^2} = {a^2} - 2ab + {b^2}$
Similarly we have the formulas including cube terms and so on.

Note: The given expression is $8 - 4a - 2{a^3} + {a^4}$. We actually took $ - {a^3}$ outside from the last two terms.
Also this problem can be factored in another way.
We can rearrange the terms such a way that $8 - 2{a^3} - 4a + {a^4}$.
Now in the same way we can take common from first two terms and last two terms.
Taking $2$ from first two terms and $a$ from last two terms we get,
$8 - 2{a^3} - 4a + {a^4} = 2(4 - {a^3}) - a(4 - {a^3})$.
Now we can see $4 - {a^3}$ is common. So proceeding we get,
$8 - 2{a^3} - 4a + {a^4} = (4 - {a^3})(2 - a)$
This gives,
$8 - 4a - 2{a^3} + {a^4} = (4 - {a^3})(2 - a)$
So we get the same factors as before.