Question

How do you factor completely: $14b{{x}^{2}}-7{{x}^{3}}-4b+2x$?

Answer 1

Hint: The polynomial in the above question, which is written as $14b{{x}^{2}}-7{{x}^{3}}-4b+2x$, is not in the standard form, that is, in the order of decreasing of powers of $x$. So firstly we need to write it in the standard form as $-7{{x}^{3}}+14b{{x}^{2}}+2x-4b$. Then we have to use the factor by grouping method to factor the given polynomial. For this, we have to group the first two and the last two terms to get $\left( -7{{x}^{3}}+14b{{x}^{2}} \right)+\left( 2x-4b \right)$. Then we have to take $-7{{x}^{2}}$ common from the first group and $2$ common from the second group. Then finally, on taking the common factor outside, the given polynomial will be factored completely.

Complete step by step solution:
Let the given polynomial be $p\left( x \right)$ so that we can write
$\Rightarrow p\left( x \right)=14b{{x}^{2}}-7{{x}^{3}}-4b+2x$
The given polynomial is not written in the standard form since the terms are not arranged in decreasing order of the powers of $x$. Therefore, we write it in the standard form as
\[\Rightarrow p\left( x \right)=-7{{x}^{3}}+14b{{x}^{2}}+2x-4b\]
Now, we group the first two and the last two terms to get
\[\Rightarrow p\left( x \right)=\left( -7{{x}^{3}}+14b{{x}^{2}} \right)+\left( 2x-4b \right)\]
As we can see that \[-7{{x}^{2}}\] is common to the first group and $2$ is common to the second group. Taking these outside of the respected groups, we get
\[\Rightarrow p\left( x \right)=-7{{x}^{2}}\left( x-2b \right)+2\left( x-2b \right)\]
Now, since the factor \[\left( x-2b \right)\] is common, we can take it outside to get
\[\Rightarrow p\left( x \right)=\left( x-2b \right)\left( -7{{x}^{2}}+2 \right)\]
Now, we take $-7$ common from the second factor to get
\[\Rightarrow p\left( x \right)=-7\left( x-2b \right)\left( {{x}^{2}}-\dfrac{2}{7} \right)\]
Putting \[\dfrac{2}{7}={{\left( \sqrt{\dfrac{2}{7}} \right)}^{2}}\] we get
\[\Rightarrow p\left( x \right)=-7\left( x-2b \right)\left( {{x}^{2}}-{{\left( \sqrt{\dfrac{2}{7}} \right)}^{2}} \right)\]
Now, using the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we can factorize further the second factor as
\[\Rightarrow p\left( x \right)=-7\left( x-2b \right)\left( x+\sqrt{\dfrac{2}{7}} \right)\left( x-\sqrt{\dfrac{2}{7}} \right)\]
Hence, the given polynomial is completely factored as \[-7\left( x-2b \right)\left( x+\sqrt{\dfrac{2}{7}} \right)\left( x-\sqrt{\dfrac{2}{7}} \right)\].

Note:
Do not end your solution at the factored form \[\left( x-2b \right)\left( -7{{x}^{2}}+2 \right)\]. Our attempt must be to factorize the given polynomial into the product of the one degree polynomials. It might be possible in some cases that the quadratic factor cannot be factored further, but if it can be factorized, then we must factorize it.