Question

How do you factor \[{f^3} + 2{f^2} - 64f - 128\] ?

Answer 1

Hint: We have a polynomial of degree 3 and it is called cubic polynomial. IF the polynomial is of degree ‘n’ then the polynomial as ‘n’ roots or ‘n’ factors. Here we will have three factors. We simplify the given cubic polynomial by taking \[{f^2}\] common in the first two terms and 64 common in the remaining two terms. After applying the suitable algebraic identities we will obtain the desired results.

Complete step by step solution:
Given,
 \[{f^3} + 2{f^2} - 64f - 128\]
Take \[{f^2}\] in the first two terms we have,
 \[ \Rightarrow {f^2}\left( {f + 2} \right) - 64f - 128\]
Now take -64 common in the remaining two term we have,
 \[ \Rightarrow {f^2}\left( {f + 2} \right) - 64\left( {f + 2} \right)\]
Now taking \[\left( {f + 2} \right)\] common we have,
 \[ \Rightarrow \left( {{f^2} - 64} \right)\left( {f + 2} \right)\]
This can rewrite it as,
 \[ \Rightarrow \left( {{f^2} - {8^2}} \right)\left( {f + 2} \right)\]
We know the algebraic identity \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\] . Applying this to the first term of the above expression we have,
 \[ \Rightarrow \left( {f + 8} \right)\left( {f - 8} \right)\left( {f + 2} \right)\] .
Thus the factors of \[{f^3} + 2{f^2} - 64f - 128\] are \[\left( {f + 8} \right)\] , \[\left( {f - 8} \right)\] and \[\left( {f + 2} \right)\] .
So, the correct answer is “ \[\left( {f + 8} \right)\] , \[\left( {f - 8} \right)\] and \[\left( {f + 2} \right)\] .”.

Note: We can also find the roots of the given cubic polynomial. All we need to do is substituting the obtained factors to zero.
 \[\left( {f + 8} \right)\left( {f - 8} \right)\left( {f + 2} \right) = 0\]
Now according to zero product principle we have,
 \[\left( {f + 8} \right) = 0\] and \[\left( {f - 8} \right) = 0\] and \[\left( {f + 2} \right) = 0\]
 \[ \Rightarrow f = - 8\] and \[f = 8\] and \[f = - 2\] . These are roots of the given polynomial. Usually we simplify the given cubic polynomial to quadratic polynomial using synthetic division and then we solve the quadratic polynomial using factorization method or quadratic polynomial.