Answer :
We want to factor completely the polynomial
[tex]$$
8x^3 + 20x^2 - 18x - 45.
$$[/tex]
A good method is to use grouping. First, group the terms into two pairs:
[tex]$$
(8x^3 + 20x^2) + (-18x - 45).
$$[/tex]
In the first group, factor out the common factor [tex]$4x^2$[/tex]:
[tex]$$
8x^3 + 20x^2 = 4x^2(2x + 5).
$$[/tex]
In the second group, factor out [tex]$-9$[/tex]:
[tex]$$
-18x - 45 = -9(2x + 5).
$$[/tex]
Now the expression becomes
[tex]$$
4x^2(2x + 5) - 9(2x + 5).
$$[/tex]
Since [tex]$(2x + 5)$[/tex] is a common factor, factor it out:
[tex]$$
(2x + 5)(4x^2 - 9).
$$[/tex]
Next, notice that the quadratic [tex]$4x^2 - 9$[/tex] is a difference of squares. Recall that the difference of squares formula is
[tex]$$
a^2 - b^2 = (a - b)(a + b).
$$[/tex]
Here, we have [tex]$a = 2x$[/tex] and [tex]$b = 3$[/tex], so
[tex]$$
4x^2 - 9 = (2x - 3)(2x + 3).
$$[/tex]
Thus, the complete factorization is
[tex]$$
(2x + 5)(2x - 3)(2x + 3).
$$[/tex]
This expression is the fully factored form of the original polynomial.
[tex]$$
8x^3 + 20x^2 - 18x - 45.
$$[/tex]
A good method is to use grouping. First, group the terms into two pairs:
[tex]$$
(8x^3 + 20x^2) + (-18x - 45).
$$[/tex]
In the first group, factor out the common factor [tex]$4x^2$[/tex]:
[tex]$$
8x^3 + 20x^2 = 4x^2(2x + 5).
$$[/tex]
In the second group, factor out [tex]$-9$[/tex]:
[tex]$$
-18x - 45 = -9(2x + 5).
$$[/tex]
Now the expression becomes
[tex]$$
4x^2(2x + 5) - 9(2x + 5).
$$[/tex]
Since [tex]$(2x + 5)$[/tex] is a common factor, factor it out:
[tex]$$
(2x + 5)(4x^2 - 9).
$$[/tex]
Next, notice that the quadratic [tex]$4x^2 - 9$[/tex] is a difference of squares. Recall that the difference of squares formula is
[tex]$$
a^2 - b^2 = (a - b)(a + b).
$$[/tex]
Here, we have [tex]$a = 2x$[/tex] and [tex]$b = 3$[/tex], so
[tex]$$
4x^2 - 9 = (2x - 3)(2x + 3).
$$[/tex]
Thus, the complete factorization is
[tex]$$
(2x + 5)(2x - 3)(2x + 3).
$$[/tex]
This expression is the fully factored form of the original polynomial.
