Answer :

Answer:

1) (x + 5)(x + 5)

2)

[tex] {(x + 1)}^{2} + {(y - 5)}^{2} = 1[/tex]

Step-by-step explanation:

Please find the attached.

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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:

[tex]\boxed{a^2 + 2ab + b^2 = (a+b)(a+b)=(a+b)^2}[/tex]

In the case of x² + 10x + 25:

  • [tex]a^2 = x^2[/tex]
  • [tex]2ab=10x[/tex]
  • [tex]b^2 = 25[/tex]

Therefore:

[tex]a^2 =x^2 \implies a = x[/tex]

[tex]b^2=25 \implies b=\sqrt{25}=\pm5[/tex]

As 2ab = 10x then b = 5:

[tex]2ab = 2 \cdot x \cdot 5 = 10x[/tex]

So, by applying the formula:

[tex]x^2 + 10x + 25 = (x + 5)(x + 5)[/tex]

Therefore, the factored polynomial is:

[tex]\Large\boxed{\boxed{(x+5)(x+5)}}[/tex]

[tex]\dotfill[/tex]

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:

[tex](x^2+2x)+(y^2-10y)=-25[/tex]

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:

[tex]\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[/tex]

Simplify:

[tex]\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[/tex]

The expressions in the brackets are perfect square trinomials. Therefore, we can factor them using the formulas:

[tex]\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}}[/tex]

So, the factored polynomial is:

[tex]\Large\boxed{\boxed{(x+1)^2+(y-5)^2=1}}[/tex]

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