Question

How do you solve \[{x^3} + 7{x^2} - x - 7 < 0?\]

Answer 1

Hint: To solve this problem we should know about inequality.
Inequality: An inequality compares two values, showing if one is less than, greater than, or simply not equal to another value.
That is $a \ne b$, $a$ is not equal to $b$ .
You should also know about:
Open bracket $ \to (\,)$
Closed bracket \[ \to [\,]\]

Complete step by step solution:
Let $f(x) = {x^3} + 7{x^2} - x - 7$
We have to find its one root by trial and error method.
Let's take \[x = 1\] . we get,
 $f(1) = 1 + 7 - 1 - 7 = 0$
So, $(x - 1)$ is a factor of $f(x)$
To find the other factors, we do a long division.
We will first multiply ${x^2}$ to divide or divide, and change the sign as per rule and do simple operation. We get,
 $
  \,\,\,\,\,\,\,\,\,\,\,{x^2} \\
  x - 1\left){\vphantom{1{{x^3} + 7{x^2} - x - 7}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{{x^3} + 7{x^2} - x - 7}}} \\
  \,\,\,\,\,\,\,\,\,\,{x^3} - {x^2} \\
  \,\,\,\,\,\,\,\,\,\,0 + 8{x^2} - x \\
  \,\,\,\,\,\,\,\,\,\, + 8{x^2} - 8x \\
 $
Then We will first multiply \[8x\] to divisor to do division, and change the sign as per rule and do simple operation. We get,
 $
  \,\,\,\,\,\,\,\,\,\,\,{x^2} + 8x \\
  x - 1\left){\vphantom{1{{x^3} + 7{x^2} - x - 7}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{{x^3} + 7{x^2} - x - 7}}} \\
  \,\,\,\,\,\,\,\,\,\,{x^3} - {x^2} \\
  \,\,\,\,\,\,\,\,\,\,0 + 8{x^2} - x \\
  \,\,\,\,\,\,\,\,\,\, + 8{x^2} - 8x \\
  \,\,\,\,\,\,\,\,\,\,\,\,\, + 0 + 7x - 7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
 $
Then We will first multiply \[7\] to divisor to do division, and change the sign as per rule and do simple operation. We get,
 $
  \,\,\,\,\,\,\,\,\,\,\,{x^2} + 8x + 7 \\
  x - 1\left){\vphantom{1{{x^3} + 7{x^2} - x - 7}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{{x^3} + 7{x^2} - x - 7}}} \\
  \,\,\,\,\,\,\,\,\,\,{x^3} - {x^2} \\
  \,\,\,\,\,\,\,\,\,\,0 + 8{x^2} - x \\
  \,\,\,\,\,\,\,\,\,\, + 8{x^2} - 8x \\
  \,\,\,\,\,\,\,\,\,\,\,\,\, + 0 + 7x - 7 \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 7x - 7 \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 0 - 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
 $
As here the remainder is zero hence the equation is completely divided by $(x - 1)$ .
Therefore, by doing factorization. We get,
 $f(x) = (x - 1)({x^2} + 8x + 7) = (x - 1)(x + 1)(x + 7)$
The value of $x$ will be zero at point:
 $(x - 1) = 0$
 $ \Rightarrow x = 1$
Or, $(x + 1)$
 $ \Rightarrow x = - 1$
And $(x + 7)$
 $ \Rightarrow x = - 7$
Now, we can build the sign chart as per above calculation:

$x$	$( - \infty , - 7)$	$( - 7, - 1)$	$( - 1,1)$	$(1, + \infty )$
$(x + 7)$	$ - $	$ + $	$ + $	$ + $
$(x + 1)$	$ - $	$ - $	$ + $	$ + $
$(x + 1)$	$ - $	$ - $	$ - $	$ + $
$f(x)$	$ - $	$ + $	$ - $	$ + $

 Therefore,
 $f(x) < 0\,when\,x \in ] - \infty , - 7[ \cup ] - 1,1[$
Hence, the answer is $\,x \in ] - \infty , - 7[ \cup ] - 1,1[$
So, the correct answer is “Option C”.

Note: This are linear inequality symbol:
 $ \leqslant \to $ less than or equal to
 $ < \to $ less than
 $ \ne \to $ not equal to
 $ \geqslant \to $ greater than or equal to
 $ > \to $ greater than