Solve left(2x+1right)^2+left(2x-1right)^2+left(2x+3right)^2=251

$${ \left(2x+1 \right) }^{ 2 } + { \left(2x-1 \right) }^{ 2 } + { \left(2x+3 \right) }^{ 2 } =251$$

Solve for x

$$x=-5$$

$$x=4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$4x^{2}+4x+1+\left(2x-1\right)^{2}+\left(2x+3\right)^{2}=251$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}+4x+1+4x^{2}-4x+1+\left(2x+3\right)^{2}=251$$

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$8x^{2}+4x+1-4x+1+\left(2x+3\right)^{2}=251$$

Combine 4x and -4x to get 0.

$$8x^{2}+1+1+\left(2x+3\right)^{2}=251$$

Add 1 and 1 to get 2.

$$8x^{2}+2+\left(2x+3\right)^{2}=251$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$8x^{2}+2+4x^{2}+12x+9=251$$

Combine 8x^{2} and 4x^{2} to get 12x^{2}.

$$12x^{2}+2+12x+9=251$$

Add 2 and 9 to get 11.

$$12x^{2}+11+12x=251$$

Subtract 251 from both sides.

$$12x^{2}+11+12x-251=0$$

Subtract 251 from 11 to get -240.

$$12x^{2}-240+12x=0$$

Divide both sides by 12.

$$x^{2}-20+x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+x-20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-20\right)=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$a=-4 b=5$$

Rewrite x^{2}+x-20 as \left(x^{2}-4x\right)+\left(5x-20\right).

$$\left(x^{2}-4x\right)+\left(5x-20\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x-4\right)+5\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+5\right)$$

To find equation solutions, solve x-4=0 and x+5=0.

$$x=4 x=-5$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$4x^{2}+4x+1+\left(2x-1\right)^{2}+\left(2x+3\right)^{2}=251$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}+4x+1+4x^{2}-4x+1+\left(2x+3\right)^{2}=251$$

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$8x^{2}+4x+1-4x+1+\left(2x+3\right)^{2}=251$$

Combine 4x and -4x to get 0.

$$8x^{2}+1+1+\left(2x+3\right)^{2}=251$$

Add 1 and 1 to get 2.

$$8x^{2}+2+\left(2x+3\right)^{2}=251$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$8x^{2}+2+4x^{2}+12x+9=251$$

Combine 8x^{2} and 4x^{2} to get 12x^{2}.

$$12x^{2}+2+12x+9=251$$

Add 2 and 9 to get 11.

$$12x^{2}+11+12x=251$$

Subtract 251 from both sides.

$$12x^{2}+11+12x-251=0$$

Subtract 251 from 11 to get -240.

$$12x^{2}-240+12x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$12x^{2}+12x-240=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 12 for b, and -240 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 12\left(-240\right)}}{2\times 12}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 12\left(-240\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-12±\sqrt{144-48\left(-240\right)}}{2\times 12}$$

Multiply -48 times -240.

$$x=\frac{-12±\sqrt{144+11520}}{2\times 12}$$

Add 144 to 11520.

$$x=\frac{-12±\sqrt{11664}}{2\times 12}$$

Take the square root of 11664.

$$x=\frac{-12±108}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-12±108}{24}$$

Now solve the equation x=\frac{-12±108}{24} when ± is plus. Add -12 to 108.

$$x=\frac{96}{24}$$

Divide 96 by 24.

$$x=4$$

Now solve the equation x=\frac{-12±108}{24} when ± is minus. Subtract 108 from -12.

$$x=-\frac{120}{24}$$

Divide -120 by 24.

$$x=-5$$

The equation is now solved.

$$x=4 x=-5$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$4x^{2}+4x+1+\left(2x-1\right)^{2}+\left(2x+3\right)^{2}=251$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}+4x+1+4x^{2}-4x+1+\left(2x+3\right)^{2}=251$$

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$8x^{2}+4x+1-4x+1+\left(2x+3\right)^{2}=251$$

Combine 4x and -4x to get 0.

$$8x^{2}+1+1+\left(2x+3\right)^{2}=251$$

Add 1 and 1 to get 2.

$$8x^{2}+2+\left(2x+3\right)^{2}=251$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$8x^{2}+2+4x^{2}+12x+9=251$$

Combine 8x^{2} and 4x^{2} to get 12x^{2}.

$$12x^{2}+2+12x+9=251$$

Add 2 and 9 to get 11.

$$12x^{2}+11+12x=251$$

Subtract 11 from both sides.

$$12x^{2}+12x=251-11$$

Subtract 11 from 251 to get 240.

$$12x^{2}+12x=240$$

Divide both sides by 12.

$$\frac{12x^{2}+12x}{12}=\frac{240}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\frac{12}{12}x=\frac{240}{12}$$

Divide 12 by 12.

$$x^{2}+x=\frac{240}{12}$$

Divide 240 by 12.

$$x^{2}+x=20$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=20+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=20+\frac{1}{4}$$

Add 20 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{81}{4}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{9}{2} x+\frac{1}{2}=-\frac{9}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=4 x=-5$$