Factor polynomial $$ 2x^{4}-7x^{3}-5x^{2}+28x-12 $$
Answer
Explanation
Step 1 :
The polynomial $ 2x^{4}-7x^{3}-5x^{2}+28x-12 $ can be factorized by using the rational root test. To apply this test first we need to find
at least one rational zero.
Polynomial roots calculator can be used to find rational zero.
In this case one rational zero is $ x = \frac{ 1 }{ 2 } $.
This means that we can divide polynomial $ 2x^{4}-7x^{3}-5x^{2}+28x-12 $ by $ 2x-1 $ (the factor theorem)
After division we have: (you can use Synthetic Division Calculator for the step-by-step explanation on how to divide polynomials. )
$$ \frac{ 2x^{4}-7x^{3}-5x^{2}+28x-12 }{ 2x-1 } = x^{3}-3x^{2}-4x+12 $$that is:
$$ 2x^{4}-7x^{3}-5x^{2}+28x-12 = ( 2x-1 )( x^{3}-3x^{2}-4x+12 ) $$Step 2 :
To factor $ x^{3}-3x^{2}-4x+12 $ we can use factoring by grouping:
Group $ \color{blue}{ x^{3} }$ with $ \color{blue}{ -3x^{2} }$ and $ \color{red}{ -4x }$ with $ \color{red}{ 12 }$ then factor each group.
$$ \begin{aligned} x^{3}-3x^{2}-4x+12 = ( \color{blue}{ x^{3}-3x^{2} } ) + ( \color{red}{ -4x+12 }) &= \\ &= \color{blue}{ x^{2}( x-3 )} + \color{red}{ -4( x-3 ) } = \\ &= (x^{2}-4)(x-3) \end{aligned} $$Step 3 :
Rewrite $ x^{2}-4 $ as:
$$ x^{2}-4 = (x)^2 - (2)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = x $ and $ II = 2 $ , we have:
$$ x^{2}-4 = (x)^2 - (2)^2 = ( x-2 ) ( x+2 ) $$