Certainly! Let's factor the polynomial [tex]\(4x^2 - 16x + 16\)[/tex] step-by-step:
1. Identify the given polynomial:
[tex]$4x^2 - 16x + 16$[/tex]
2. Rewrite the polynomial in a standard form to look for a pattern:
Firstly, observe that this is a quadratic polynomial. We will check if it can be factored as a perfect square trinomial. A perfect square trinomial has the form [tex]\((ax - b)^2 = a^2x^2 - 2abx + b^2\)[/tex].
3. Comparison with the standard form:
[tex]$a^2x^2 - 2abx + b^2 \quad \text{with} \quad 4x^2 - 16x + 16$[/tex]
From this comparison:
- [tex]\(a^2x^2 = 4x^2 \implies a^2 = 4 \implies a = 2\)[/tex]
- [tex]\(b^2 = 16 \implies b = 4\)[/tex]
- Check the middle term:
The middle term, as given, is [tex]\(-16x\)[/tex]. We need to verify:
[tex]$-2abx = -2 \times 2 \times 4 \times x = -16x$[/tex]
This matches the given polynomial.
4. Rewrite the polynomial as a perfect square:
Based on the above observations, we can rewrite the polynomial [tex]\(4x^2 - 16x + 16\)[/tex] as:
[tex]$(2x - 4)^2$[/tex]
5. Verification:
To ensure our factorization is correct, let's expand [tex]\((2x - 4)^2\)[/tex]:
[tex]$ (2x - 4)(2x - 4) = 4x^2 - 8x - 8x + 16 = 4x^2 - 16x + 16 $[/tex]
This confirms that our factorization is accurate.
Thus, the completely factored form of the polynomial [tex]\(4x^2 - 16x + 16\)[/tex] is:
[tex]$ (2x - 4)^2 $[/tex]
