Question

Simplify the expression $({x^2} + 3)(x - 3) + 9$.

Answer 1

The FOIL method gives us the method for multiplying two binomials:
$(a + b)(c + d) = ac + ad + bc + bd$.
We will use the FOIL method to simplify the term $({x^2} + 3)(x - 3)$ first.
Then, we will substitute the simplified value of the above-mentioned term in the given expression $({x^2} + 3)(x - 3) + 9$ to find out the final simplified value of the given expression.

Complete step by step solution:
Firstly, we are simplifying the term $({x^2} + 3)(x - 3)$ using the FOIL method.
According to the FOIL method, we have that any two binomials $(a + b)$ and $(c + d)$ can be multiplied as follows:
$(a + b)(c + d) = ac + ad + bc + bd$.
So, putting $a = {x^2}$,
$b = 3$,
$c = x$ and
$d = - 3$,
into the FOIL method, we get:
$({x^2} + 3)(x - 3) = ({x^2} \times x) + ({x^2} \times ( - 3)) + (3 \times x) + (3 \times ( - 3))$.
The first bracket on the right- hand side of the above equation can be simplified using the following rule for exponents:
${a^m} \times {a^n} = {a^{m + n}}$.
The second and the fourth bracket can be simplified using the following property:
$a \times ( - b) = - ab$.
Now, multiplying the terms in the brackets, we get:
$({x^2} + 3)(x - 3) = {x^3} - 3{x^2} + 3x - 9$
Now, substitute the above simplified value of the term $({x^2} + 3)(x - 3)$ into the expression $({x^2} + 3)(x - 3) + 9$, we get:
$({x^2} + 3)(x - 3) + 9 = ({x^3} - 3{x^2} + 3x - 9) + 9$.
Now, in the above expression there are two terms which are the same but they are with different signs, that is, we have two $9$’s in the above expression but both are with opposite signs, so we can cancel them out to obtain the final simplified value of the given expression.
$\therefore$ The simplified value of the given expression is $({x^2} + 3)(x - 3) + 9 = {x^3} - 3{x^2} + 3x$.

Note:
A binomial is a polynomial that has two terms.
For example, $x + 3$, ${x^6} + 9$ are binomials as they contain only two terms.
So, be careful while using the FOIL method as it cannot be used to multiply any two polynomials. It can only be used to multiply two binomials.