Square of sum: More than one factorization

Let's solve the equation. First, we'll simplify the algebraic expressions using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$We'll apply the mentioned formula and expand the parentheses in the expressions in the equation:

$(x+1)^2=x^2 \\ x^2+2\cdot x\cdot1+1^2=x^2 \\ x^2+2x+1=x^2 \\$We'll continue and combine like terms, by moving terms between sides. Later - we can notice that the squared term cancels out, therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^2+2x+1=x^2 \\ 2x=-1\hspace{8pt}\text{/}:2\\ \boxed{x=-\frac{1}{2}}$Therefore, the correct answer is answer A.

Let's solve the equation. First, we'll simplify the algebraic expressions using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$We'll apply the mentioned formula and expand the parentheses in the expressions in the equation:

$(x+2)^2-12=x^2 \\ x^2+2\cdot x\cdot2+2^2-12=x^2 \\ x^2+4x+4-12=x^2 \\$We'll continue and combine like terms, by moving terms between sides. Then we can notice that the squared term cancels out, therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^2+4x+4-12=x^2 \\ 4x=8\hspace{8pt}\text{/}:4\\ \boxed{x=2}$Therefore, the correct answer is answer B.

Solve the equation by simplifying the expression on the left side in two stages. First, we'll multiply the expressions within the two leftmost pairs of parentheses:

Apply the shortened multiplication formula for squaring a binomial:

$(a\pm b)^2=a^2\pm2ab+b^2$

Given that these two pairs of parentheses are being multiplied by another expression (which is also in parentheses), we'll place the result inside of parentheses (marked with an underline):

$\underline{ (x-1)(x+1)}(x+1)=x^2+x^3 \\ \underline{ (x^2-1^2)}(x+1)=x^2+x^3 \\ (x^2-1)(x+1)=x^2+x^3$

Continue to simplify the expression on the left side by using the expanded distribution law:

$(a+b)(c+d)=ac+ad+bc+bd$

Additionally, we'll apply the law of exponents for multiplying terms with equal bases:

$a^ma^n=a^{m+n}$

Apply these laws in order to expand the parentheses in the expression in the equation:

$(x^2-1)(x+1)=x^2+x^3 \\ x^3+x^2-x-1=x^2+x^3 \\$Continue to combine like terms, while moving terms between sides. Later - we observe that the terms with squared and cubed powers cancel out, therefore it's a first-degree equation, which we'll solve by isolating the variable term and dividing both sides of the equation by its coefficient:

$x^3+x^2-x-1=x^2+x^3 \\ -x=1\hspace{8pt}\text{/}:(-1)\\ \boxed{x=-1}$

Therefore, the correct answer is answer A.

Solve the following equation. First, we'll simplify the algebraic expressions using the square of binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$

Apply the mentioned formula and expand the parentheses in the expressions in the equation. On the right side, since we have parentheses with an exponent multiplier, we'll expand the (existing) parentheses using the square of binomial formula into additional parentheses (marked with an underline in the following calculation):

$x^2+(x-2)^2=2\underline{(x+1)^2} \\ x^2+x^2-2\cdot x\cdot2+2^2=2\underline{(x^2+2\cdot x\cdot1+1^2)} \\ x^2+x^2-4x+4=2(x^2+2x+1) \\ x^2+x^2-4x+4=2x^2+4x+2 \\$In the final stage, we expand the parentheses on the right side by using the distributive property,

Continue to combine like terms, by moving terms between sides. We observe that the squared term cancels out, therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^2+x^2-4x+4=2x^2+4x+2 \\ -8x=-2\hspace{8pt}\text{/}:(-8)\\ \boxed{x=\frac{2}{8}=\frac{1}{4}}$

In the final stage, we simplified the fraction that was obtained as the solution for $x$.

Therefore, the correct answer is answer B.

Let's solve the equation. First, we'll simplify the algebraic expressions using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$We'll apply this formula and expand the parentheses in the expressions in the equation:

$(x+3)^2=(x-3)^2 \\ x^2+2\cdot x\cdot3+3^2=x^2-2\cdot x\cdot3+3^2 \\ x^2+6x+9=x^2-6x+9 \\$We'll continue and combine like terms, by moving terms between sides. Then we can notice that the squared term cancels out, therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^2+6x+9=x^2-6x+9 \\ 12x=0\hspace{8pt}\text{/}:12\\ \boxed{x=0}$Therefore, the correct answer is answer A.