Factor polynomial $$ x^{4}-x^{3}-42x^{2} $$

$$ x^{4}-x^{3}-42x^{2} = x^{2}( x+6 )( x-7 ) $$

Step 1 :

After factoring out $ x^{2} $ we have:

$$ x^{4}-x^{3}-42x^{2} = x^{2} ( x^{2}-x-42 ) $$

Step 2 :

Step 2: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = -1 } ~ \text{ and } ~ \color{red}{ c = -42 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -1 } $ and multiply to $ \color{red}{ -42 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -42 }$.

PRODUCT = -42
-1     42	1     -42
-2     21	2     -21
-3     14	3     -14
-6     7	6     -7

Step 4: Find out which pair sums up to $\color{blue}{ b = -1 }$

PRODUCT = -42 and SUM = -1
-1     42	1     -42
-2     21	2     -21
-3     14	3     -14
-6     7	6     -7

Step 5: Put 6 and -7 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-x-42 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-x-42 & = (x + 6)(x -7) \end{aligned} $$

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