Question

How do you factor $16{x^3} + 54$ ?

Answer 1

Hint: In the given function, if we common out a factor from both terms, then we can get an equation of the form ${a^3} + {b^3}$ . To factorize ${a^3} + {b^3}$ , we have a postulate which states that ${a^3} + {b^3}$ always has a factor of $a + b$ . With the help of this one factor, we can find the other factors.

Complete step-by-step solution:
From the given equation, considering the coefficient of the first term, we can factorize it as follows
$\Rightarrow 16 = 8 \times 2$
Here, we consider such factors because $\;8$ is a perfect cube of $\;2$
Similarly, for the second term, we can factorize it as follows
$\Rightarrow 54 = 27 \times 2$
Where $\;27$ is a perfect cube of $\;3$ and has a common factor $\;2$ with the coefficient of the first term.
Rewriting the equation with the common factor
$\Rightarrow (8 \times 2){x^3} + (27 \times 2)$
$\Rightarrow 2 \times (8{x^3} + 27)$
Rewriting the equation in the perfect cube form
$\Rightarrow 2({2^3}{x^3} + {3^3})$
$\Rightarrow 2[{(2x)^3} + {3^3}]$
Comparing the equation with the standard equation ${a^3} + {b^3}$ , we get
$a = 2x$ and $b = 3$
Now, we have a postulate, which says that for every ${a^n} + {b^n}$ , where $n = odd$ , $a + b$ is always one of the factors.
Here, $n = 3$ which is $\;odd$ , hence we can say that one of the factors is $a + b$
Substituting the value of $a$ and $b$ , we can say $2x + 3$ is one of the factors of the equation
As the highest power of $x$ in this factor is $\;1$ , we need another factor with the highest power $\;2$ so that by combining the terms we can get a term with the power of $\;3$
For the highest power $\;2$ , we need a quadratic equation.
Hence, the other factor of the equation is a quadratic equation of the form $a{x^2} + bx + c$ .
We can write the factorized equation as
$\Rightarrow 2[{(2x)^3} + {3^3}] = 2[(2x + 3)(a{x^2} + bx + c)]$
If we consider $2x + 3$ as a single value, the distributive law can be applied as follows
$\Rightarrow 2[{(2x)^3} + {3^3}] = 2[a{x^2}(2x + 3) + bx(2x + 3) + c(2x + 3)]$
Similarly, applying the distributive law again,
$\Rightarrow 2[{(2x)^3} + {3^3}] = 2[2a{x^3} + 3a{x^2} + 2b{x^2} + 3bx + 2cx + 3c]$
$\Rightarrow 2[8{x^3} + 27] = 2[2a{x^3} + 3a{x^2} + 2b{x^2} + 3bx + 2cx + 3c]$
The left-hand side of the equation can be written as,
$\Rightarrow 2[8{x^3} + 0{x^2} + 0x + 27] = 2[2a{x^3} + (3a{x^2} + 2b{x^2}) + (3bx + 2cx) + 3c]$
Now, we can compare the coefficients of the terms with the same power as follows,
$2a = 8$ , $3a + 2b = 0$ , $3b + 2c = 0$ , and $3c = 27$
 Now, for the first equation, dividing both sides by $\;2$
$\Rightarrow a = \dfrac{8}{2} = 4$
Similarly, for the last equation, dividing both sides by $\;3$
$\Rightarrow c = \dfrac{{27}}{3} = 9$
Considering the second equation, and substituting $a = 4$ ,
$\Rightarrow 3(4) + 2b = 0$
$\Rightarrow 12 + 2b = 0$
Subtracting $\;12$ from both sides of the equation,
$\Rightarrow 12 + 2b - 12 = - 12$
$\Rightarrow 2b = - 12$
Dividing both sides of the equation by $\;2$ ,
$\Rightarrow b = \dfrac{{ - 12}}{2}$
$\Rightarrow b = - 6$
Now, substituting all the values obtained in the factored equation,
$\Rightarrow 2[8{x^3} + 27] = 2[(2x + 3)(4{x^2} - 6x + 9)]$
Hence the given equation can be factorized as,
$\Rightarrow 16{x^3} + 54 = (2)(2x + 3)(4{x^2} - 6x + 9)$

Therefore the value of the given expression is $(2)(2x + 3)(4{x^2} - 6x + 9)$.

Note: This postulate given here can be proved by the principle of mathematical induction. Here, we can also directly use the proven formula ${a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$ to factorize the given equation, but mistakes might occur if one doesn’t remember the equation perfectly.