Question

The product of three consecutive odd integers is $ - 6783$. How do you write and solve an equation to find the numbers?

Answer 1

Hint: We are given the product of two numbers. We have to find the numbers. First, we will assume the first number and determine two other consecutive odd numbers. Then, write the product of the numbers and equate it to the value of the product. Then, solve the expression at the left hand side to form an equation. Then, solve the equation to find the number. Then, we will determine the other two consecutive numbers.

Complete answer:
We will start by assuming the number be $x$
Now, we will determine the other two consecutive odd numbers by adding 2 and 4 to the numbers.
$ \Rightarrow \left( {x + 2} \right)$ and $ \left( {x + 4} \right)$
Now, the product of these three numbers is $ - 6783$.
$ \Rightarrow x\left( {x + 2} \right)\left( {x + 4} \right) = - 6783$
Multiply the first two expressions at the left hand side of the equation.
$ \Rightarrow \left( {{x^2} + 2x} \right)\left( {x + 4} \right) = - 6783$
Now, apply the distributive property at the left hand side of the expression.
$ \Rightarrow {x^3} + 4{x^2} + 2{x^2} + 8x = - 6783$
Combine like terms, we get:
$ \Rightarrow {x^3} + 6{x^2} + 8x = - 6783$
Add 6783 to both sides to rewrite the equation in standard form.
$ \Rightarrow {x^3} + 6{x^2} + 8x + 6783 = 0$
Here in the polynomial, $q = 1$ and $p = 6783$. First, we will find the factors of the trailing constant.
$p = 1,3,7,17,19,21,51,57,119,133$
Therefore, the possible rational roots of the polynomial are $ \pm \dfrac{1}{1}, \pm \dfrac{3}{1}, \pm \dfrac{7}{1}, \pm \dfrac{{17}}{1}, \pm \dfrac{{19}}{1}, \pm \dfrac{{21}}{1}, \pm \dfrac{{51}}{1}, \pm \dfrac{{57}}{1}, \pm \dfrac{{119}}{1}, \pm \dfrac{{133}}{1}$
Now we will substitute the value of roots one by one into the polynomial to check for which value the polynomial is zero.
By substituting $x = - 1$ into the polynomial.
$ \Rightarrow {\left( { - 1} \right)^3} + 6{\left( { - 1} \right)^2} + 8\left( { - 1} \right) + 6783$
$ \Rightarrow - 1 + 6 - 8 + 6783$
$ \Rightarrow 6780$
By substituting $x = - 21$into the polynomial.
$ \Rightarrow {\left( { - 21} \right)^3} + 6{\left( { - 21} \right)^2} + 8\left( { - 21} \right) + 6783$
$ \Rightarrow - 9261 + 2646 - 168 + 6783 = 0$
Thus, $x = - 21$ is a root of the polynomial which means $x + 21$ is a factor of the polynomial.
We will find other factors using the long division method.
$x + 21)\overline {{x^3} + 6{x^2} + 8x + 6783} ({x^2} - 15x + 323$
             $\underline {{x^3} + 21{x^2}} $
                  $ - 15{x^2} + 8x$
                     $\underline {15{x^2} - 315x} $
                                  $323x + 6783$
                                  $\underline {323x + 6783} $
                                                   $0$
Thus, the factors of the polynomial are $\left( {x + 21} \right)\left( {{x^2} - 15x + 323} \right)$
Now, factorize the polynomial ${x^2} - 15x + 323$ by substituting $a = 1$, $b = - 15$ and $c = 323$ into the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$ \Rightarrow x = \dfrac{{ - \left( { - 15} \right) \pm \sqrt {{{\left( { - 15} \right)}^2} - 4\left( 1 \right)\left( {323} \right)} }}{{2\left( 1 \right)}}$
$ \Rightarrow x = \dfrac{{15 \pm \sqrt {225 - 1292} }}{2}$
$ \Rightarrow x = \dfrac{{15 \pm \sqrt { - 1067} }}{2}$
Since the discriminant of the equation is negative, which means the equation has no real roots.
Therefore, the only root of the equation is $x = - 21$
Now, determine the other two consecutive odd integers.
$ \Rightarrow \left( { - 21 + 2} \right) = - 19$ and
$ \Rightarrow \left( { - 21 + 4} \right) = - 17$

Hence, the three consecutive odd integers are $ - 21$, $ - 19$ and $ - 17$

Note: The students must note that in such types of questions, to factor the polynomial synthetic division can also be used instead of the long division process. The main approach to solve this question is to form a question in such a way and then apply identity.