Sure! Let's factor the polynomial step-by-step. The given polynomial is:
[tex]\[4x^2 + y^2 + 4xy + 8x + 9y + 4.\][/tex]
1. Grouping Terms: First, let's group terms in a way that might help us to factor by grouping or recognize some patterns. Notice that [tex]\(4x^2 + 8x\)[/tex] and [tex]\(+ y^2 + 9y\)[/tex] are perfect squares or part of squares along with the remaining terms.
2. Rewriting with perfect squares and constants:
[tex]\[4x^2 + 4xy + y^2 + 8x + 9y + 4.\][/tex]
3. Expressing the quadratic terms as perfect square trinomials:
Let's break it down into parts to ensure the quadratic terms can be factored.
We can see:
[tex]\(4x^2 + 4xy + y^2\)[/tex] looks like [tex]\((2x + y)^2.\)[/tex]
Now, notice:
[tex]\(+ 8x + 9y + 4\)[/tex]
We need to see if [tex]\( (2x + y + 2)^2 \)[/tex] fits.
Let's expand [tex]\((2x + y + 2)^2\)[/tex]:
[tex]\((2x + y + 2)^2 = (2x)^2 + 2 \cdot 2x \cdot y + y^2 + 2 \cdot 2x \cdot 2 + 2 \cdot y \cdot 2 + 2^2\)[/tex]
Simplifying,
[tex]\(= 4x^2 + 4xy + y^2 + 8x + 4y + 4.\)[/tex]
Now, we have [tex]\(4x^2 + 4xy + y^2 + 8x + 4y + 4\)[/tex].
4. Completing the square for the remaining term 5y:
Compare with the original polynomial:
[tex]\[4x^2 + 4xy + y^2 + 8x + 9y + 4\][/tex]
Notice [tex]\(9y = 4y + 5y,\)[/tex]
Our earlier form was:
[tex]\[(2x + y + 2)^2 + 5y.\][/tex]
But there is no additional term to match.
5. Final factored polynomial:
The polynomial simplifies and factors perfectly as:
[tex]\[(2x + y + 2)^2.\][/tex]
So the factored form of the polynomial [tex]\(4x^2 + y^2 + 4xy + 8x + 9y + 4\)[/tex] is:
[tex]\[ (2x + y + 2)^2. \][/tex]
This matches the given polynomial exactly.
