Question

How do you factor ${(x - 3)^3} + {(3x - 2)^3}$?

Answer 1

Hint: The given equation is ${(x - 3)^3} + {(3x - 2)^3}$. Both terms are cubes and expression can be factored as
${a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$
For factoring polynomials, “factoring” is always done using some set of numbers as a possible coefficient.
We use the square and cubic formula.

Complete step-by-step solution:
The given equation is ${(x - 3)^3} + {(3x - 2)^3}$
Both terms are cubes and expression can be factored as
${a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$
To make the process easier to follow,
Let $(x - 3) = m$ and $(3x - 2) = n$
Hence, ${m^3} + {n^3} = (m + n)({m^2} - mn + {n^2})$
We substitute the $m$ and $n$ in the expansion of ${m^3} + {n^3}$
${m^3} + {n^3} = (x - 3 + 3x - 2)({(x - 3)^2} - (x - 3)(3x - 2) + {(3x - 2)^2})$
Now we simplify
First we add the variable and constant in the first term, hence we get
$ = (4x - 5)({(x - 3)^2} - (x - 3)(3x - 2) + {(3x - 2)^2})$
Now the second term has there are two square expressions, so we first simplify that the expression is\[{(x - 3)^2}\],
Here,${a^2} - {b^2} = {a^2} - 2ab + {b^2}$ , hence
$\Rightarrow$${(x - 3)^2} = {x^2} - 2(x)(3) + {3^2}$
Now multiply $2$ by $3$ and square the last term, hence we get
$\Rightarrow$${(x - 3)^2} = {x^2} - 6x + 9$
The next expression is ${(3x - 2)^2}$
$\Rightarrow$${(3x - 2)^2} = {(3x)^2} - 2(3x)(2) + {2^2}$
Multiply $2$ by $3$
$\Rightarrow$${(3x - 2)^2} = {(3x)^2} - 6x(2) + {2^2}$
Multiply $2$ by $6$
$\Rightarrow$${(3x - 2)^2} = {(3x)^2} - 12x + {2^2}$
Now square the first and the last term
$\Rightarrow$${(3x - 2)^2} = 9{x^2} - 12x + 4$
Now we substitute $(x - 3)$ and $(3x - 2)$ in the expansion
$ = (4x - 5)(({x^2} - 6x + 9) - (x - 3)(3x - 2) + (9{x^2} - 12x + 4))$
Expand this term $(x - 3)(3x - 2)$
Multiply each term
$\Rightarrow$$3{x^2} - 2x - 9x + 6$
Add the $x$ term
$\Rightarrow$$3{x^2} - 11x + 6$
Substitute in the expansion
\[ = (4x - 5)(({x^2} - 6x + 9) - (3{x^2} - 11x + 6) + (9{x^2} - 12x + 4))\]
Open the brackets
\[ = (4x - 5)({x^2} - 6x + 9 - 3{x^2} + 11x - 6 + 9{x^2} - 12x + 4)\]
Now we add and subtract in the expansion
\[ = (4x - 5)(7{x^2} - 7x + 7)\]
We take common by $7$
\[ = 7(4x - 5)({x^2} - x + 1)\]

Therefore ${(x - 3)^3} + {(3x - 2)^3} = 7(4x - 5)({x^2} - x + 1)$

Note: A polynomial is an algebraic expression with one or more terms in which a constant and a variable are separated by an addition or a subtraction sign. The general form of a polynomial is \[a{x^n} + b{x^{n - 1}} + \ldots + kx + l\], where each variable has a constant accompanying its coefficient. Now that you have an understanding of how to use the remainder theorem to find the remainder of polynomials without actual division, the next theorem to look at is called the Factor theorem. In mathematics, a factor is a number of expressions that divides another number or expressions to get a whole number with no remainder.