Solve [-(2m+1)]^2-4(m^2-2)+m^2=21

$$[ - ( 2 m + 1 ) ] ^ { 2 } - 4 ( m ^ { 2 } - 2 ) + m ^ { 2 } = 21$$

Solve for m

$$m=-6$$

$$m=2$$

Steps Using Factoring

To find the opposite of 2m+1, find the opposite of each term.

$$\left(-2m-1\right)^{2}-4\left(m^{2}-2\right)+m^{2}=21$$

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

$$4m^{2}+4m+1-4\left(m^{2}-2\right)+m^{2}=21$$

Use the distributive property to multiply -4 by m^{2}-2.

$$4m^{2}+4m+1-4m^{2}+8+m^{2}=21$$

Combine 4m^{2} and -4m^{2} to get 0.

$$4m+1+8+m^{2}=21$$

Add 1 and 8 to get 9.

$$4m+9+m^{2}=21$$

Subtract 21 from both sides.

$$4m+9+m^{2}-21=0$$

Subtract 21 from 9 to get -12.

$$4m-12+m^{2}=0$$

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

$$m^{2}+4m-12=0$$

To solve the equation, factor m^{2}+4m-12 using formula m^{2}+\left(a+b\right)m+ab=\left(m+a\right)\left(m+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=-2 b=6$$

Rewrite factored expression \left(m+a\right)\left(m+b\right) using the obtained values.

$$\left(m-2\right)\left(m+6\right)$$

To find equation solutions, solve m-2=0 and m+6=0.

$$m=2 m=-6$$

Steps Using Factoring By Grouping

To find the opposite of 2m+1, find the opposite of each term.

$$\left(-2m-1\right)^{2}-4\left(m^{2}-2\right)+m^{2}=21$$

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

$$4m^{2}+4m+1-4\left(m^{2}-2\right)+m^{2}=21$$

Use the distributive property to multiply -4 by m^{2}-2.

$$4m^{2}+4m+1-4m^{2}+8+m^{2}=21$$

Combine 4m^{2} and -4m^{2} to get 0.

$$4m+1+8+m^{2}=21$$

Add 1 and 8 to get 9.

$$4m+9+m^{2}=21$$

Subtract 21 from both sides.

$$4m+9+m^{2}-21=0$$

Subtract 21 from 9 to get -12.

$$4m-12+m^{2}=0$$

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

$$m^{2}+4m-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as m^{2}+am+bm-12. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-12\right)=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=-2 b=6$$

Rewrite m^{2}+4m-12 as \left(m^{2}-2m\right)+\left(6m-12\right).

$$\left(m^{2}-2m\right)+\left(6m-12\right)$$

Factor out m in the first and 6 in the second group.

$$m\left(m-2\right)+6\left(m-2\right)$$

Factor out common term m-2 by using distributive property.

$$\left(m-2\right)\left(m+6\right)$$

To find equation solutions, solve m-2=0 and m+6=0.

$$m=2 m=-6$$

Steps Using the Quadratic Formula

To find the opposite of 2m+1, find the opposite of each term.

$$\left(-2m-1\right)^{2}-4\left(m^{2}-2\right)+m^{2}=21$$

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

$$4m^{2}+4m+1-4\left(m^{2}-2\right)+m^{2}=21$$

Use the distributive property to multiply -4 by m^{2}-2.

$$4m^{2}+4m+1-4m^{2}+8+m^{2}=21$$

Combine 4m^{2} and -4m^{2} to get 0.

$$4m+1+8+m^{2}=21$$

Add 1 and 8 to get 9.

$$4m+9+m^{2}=21$$

Subtract 21 from both sides.

$$4m+9+m^{2}-21=0$$

Subtract 21 from 9 to get -12.

$$4m-12+m^{2}=0$$

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.

$$m^{2}+4m-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 4.

$$m=\frac{-4±\sqrt{16-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$m=\frac{-4±\sqrt{16+48}}{2}$$

Add 16 to 48.

$$m=\frac{-4±\sqrt{64}}{2}$$

Take the square root of 64.

$$m=\frac{-4±8}{2}$$

Now solve the equation m=\frac{-4±8}{2} when ± is plus. Add -4 to 8.

$$m=\frac{4}{2}$$

Divide 4 by 2.

$$m=2$$

Now solve the equation m=\frac{-4±8}{2} when ± is minus. Subtract 8 from -4.

$$m=-\frac{12}{2}$$

Divide -12 by 2.

$$m=-6$$

The equation is now solved.

$$m=2 m=-6$$

Steps for Completing the Square

To find the opposite of 2m+1, find the opposite of each term.

$$\left(-2m-1\right)^{2}-4\left(m^{2}-2\right)+m^{2}=21$$

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

$$4m^{2}+4m+1-4\left(m^{2}-2\right)+m^{2}=21$$

Use the distributive property to multiply -4 by m^{2}-2.

$$4m^{2}+4m+1-4m^{2}+8+m^{2}=21$$

Combine 4m^{2} and -4m^{2} to get 0.

$$4m+1+8+m^{2}=21$$

Add 1 and 8 to get 9.

$$4m+9+m^{2}=21$$

Subtract 9 from both sides.

$$4m+m^{2}=21-9$$

Subtract 9 from 21 to get 12.

$$4m+m^{2}=12$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$m^{2}+4m=12$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}+4m+2^{2}=12+2^{2}$$

Square 2.

$$m^{2}+4m+4=12+4$$

Add 12 to 4.

$$m^{2}+4m+4=16$$

Factor m^{2}+4m+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}.

$$\left(m+2\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m+2\right)^{2}}=\sqrt{16}$$

Simplify.

$$m+2=4 m+2=-4$$

Subtract 2 from both sides of the equation.

$$m=2 m=-6$$