Answer :
To factor the polynomial
[tex]$$27x^2 + 18x + 3,$$[/tex]
follow these steps:
1. Factor out the greatest common factor (GCF):
Notice that each term in the polynomial is divisible by 3. Factor 3 out:
[tex]$$27x^2 + 18x + 3 = 3(9x^2 + 6x + 1).$$[/tex]
2. Factor the quadratic inside the parentheses:
Now, focus on the quadratic
[tex]$$9x^2 + 6x + 1.$$[/tex]
We look for a product of two binomials. Notice that if we consider the square of the binomial [tex]$(3x + 1)$[/tex], we have:
[tex]$$(3x+1)^2 = (3x+1)(3x+1).$$[/tex]
Expanding [tex]$(3x+1)^2$[/tex] gives:
[tex]$$ (3x+1)^2 = 9x^2 + 2\cdot 3x \cdot 1 + 1^2 = 9x^2 + 6x + 1.$$[/tex]
This matches the quadratic inside the parentheses.
3. Write the complete factorization:
Substitute the factored quadratic back:
[tex]$$27x^2 + 18x + 3 = 3(3x+1)^2.$$[/tex]
Thus, the completely factored form is:
[tex]$$\boxed{3(3x+1)^2}.$$[/tex]
[tex]$$27x^2 + 18x + 3,$$[/tex]
follow these steps:
1. Factor out the greatest common factor (GCF):
Notice that each term in the polynomial is divisible by 3. Factor 3 out:
[tex]$$27x^2 + 18x + 3 = 3(9x^2 + 6x + 1).$$[/tex]
2. Factor the quadratic inside the parentheses:
Now, focus on the quadratic
[tex]$$9x^2 + 6x + 1.$$[/tex]
We look for a product of two binomials. Notice that if we consider the square of the binomial [tex]$(3x + 1)$[/tex], we have:
[tex]$$(3x+1)^2 = (3x+1)(3x+1).$$[/tex]
Expanding [tex]$(3x+1)^2$[/tex] gives:
[tex]$$ (3x+1)^2 = 9x^2 + 2\cdot 3x \cdot 1 + 1^2 = 9x^2 + 6x + 1.$$[/tex]
This matches the quadratic inside the parentheses.
3. Write the complete factorization:
Substitute the factored quadratic back:
[tex]$$27x^2 + 18x + 3 = 3(3x+1)^2.$$[/tex]
Thus, the completely factored form is:
[tex]$$\boxed{3(3x+1)^2}.$$[/tex]
