Answer:
1) (x + 5)(x + 5)
2) (x + 1)² + (y - 5)² = 1
Step-by-step explanation:
Question 1
A polynomial in the form of a perfect square trinomial a² + 2ab + b² can be factored into the square of a binomial:
\(\boxed{a^2 + 2ab + b^2 = (a+b)(a+b)=(a+b)^2}\)
In the case of x² + 10x + 25:
- \(a^2 = x^2\)
- \(2ab=10x\)
- \(b^2 = 25\)
Therefore:
\(a^2 =x^2 \implies a = x\)
\(b^2=25 \implies b=\sqrt{25}=\pm5\)
As 2ab = 10x then b = 5:
\(2ab = 2 \cdot x \cdot 5 = 10x\)
So, by applying the formula:
\(x^2 + 10x + 25 = (x + 5)(x + 5)\)
Therefore, the factored polynomial is:
\(\Large\boxed{\boxed{(x+5)(x+5)}}\)
\(\dotfill\)
Question 2
To write x² + y² + 2x - 10y + 25 = 0 in standard form, we need to complete the square for both the x and y terms.
Begin by grouping the terms with the same variable together and moving the constant to the right side of the equation:
\((x^2+2x)+(y^2-10y)=-25\)
Now, add the square of half the coefficient of the x-term and the square of half the coefficient of the y-term to both sides of the equation:
\(\left(x^2+2x+\left(\dfrac{2}{2}\right)^2\right)+\left(y^2-10y+\left(\dfrac{10}{2}\right)^2\right)=-25+\left(\dfrac{2}{2}\right)^2+\left(\dfrac{10}{2}\right)^2\)
Simplify:
\(\left(x^2+2x+1^2\right)+\left(y^2-10y+5^2\right)=-25+1^2+5^2\\\\\\\left(x^2+2x+1^2\right)+\left(y^2-10y+5^2\right)=-25+1+25\\\\\\\left(x^2+2x+1^2\right)+\left(y^2-10y+5^2\right)=1\)
The expressions in the brackets are perfect square trinomials. Therefore, we can factor them using the formulas:
\(\boxed{\begin{array}{l}a^2 + 2ab + b^2 = (a+b)(a+b)=(a+b)^2\\\\a^2-2ab+b^2=(a-b)(a-b)=(a-b)^2\end{array}}\)
So, the factored polynomial is:
\(\Large\boxed{\boxed{(x+1)^2+(y-5)^2=1}}\)
