Solve (1+2z)^3-(z+2)(z^2-2z+4)=3(3/2+2z)+7(z^3-1)

$$( 1 + 2 z ) ^ { 3 } - ( z + 2 ) ( z ^ { 2 } - 2 z + 4 ) = 3 ( \frac { 3 } { 2 } + 2 z ) + 7 ( z ^ { 3 } - 1 )$$

Solve for z

$$z=\frac{\sqrt{6}}{4}\approx 0.612372436$$

$$z=-\frac{\sqrt{6}}{4}\approx -0.612372436$$

Steps by Finding Square Root

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(1+2z\right)^{3}.

$$1+6z+12z^{2}+8z^{3}-\left(z+2\right)\left(z^{2}-2z+4\right)=3\left(\frac{3}{2}+2z\right)+7\left(z^{3}-1\right)$$

Use the distributive property to multiply z+2 by z^{2}-2z+4 and combine like terms.

$$1+6z+12z^{2}+8z^{3}-\left(z^{3}+8\right)=3\left(\frac{3}{2}+2z\right)+7\left(z^{3}-1\right)$$

To find the opposite of z^{3}+8, find the opposite of each term.

$$1+6z+12z^{2}+8z^{3}-z^{3}-8=3\left(\frac{3}{2}+2z\right)+7\left(z^{3}-1\right)$$

Combine 8z^{3} and -z^{3} to get 7z^{3}.

$$1+6z+12z^{2}+7z^{3}-8=3\left(\frac{3}{2}+2z\right)+7\left(z^{3}-1\right)$$

Subtract 8 from 1 to get -7.

$$-7+6z+12z^{2}+7z^{3}=3\left(\frac{3}{2}+2z\right)+7\left(z^{3}-1\right)$$

Use the distributive property to multiply 3 by \frac{3}{2}+2z.

$$-7+6z+12z^{2}+7z^{3}=\frac{9}{2}+6z+7\left(z^{3}-1\right)$$

Use the distributive property to multiply 7 by z^{3}-1.

$$-7+6z+12z^{2}+7z^{3}=\frac{9}{2}+6z+7z^{3}-7$$

Subtract 7 from \frac{9}{2} to get -\frac{5}{2}.

$$-7+6z+12z^{2}+7z^{3}=-\frac{5}{2}+6z+7z^{3}$$

Subtract 6z from both sides.

$$-7+6z+12z^{2}+7z^{3}-6z=-\frac{5}{2}+7z^{3}$$

Combine 6z and -6z to get 0.

$$-7+12z^{2}+7z^{3}=-\frac{5}{2}+7z^{3}$$

Subtract 7z^{3} from both sides.

$$-7+12z^{2}+7z^{3}-7z^{3}=-\frac{5}{2}$$

Combine 7z^{3} and -7z^{3} to get 0.

$$-7+12z^{2}=-\frac{5}{2}$$

Add 7 to both sides.

$$12z^{2}=-\frac{5}{2}+7$$

Add -\frac{5}{2} and 7 to get \frac{9}{2}.

$$12z^{2}=\frac{9}{2}$$

Divide both sides by 12.

$$z^{2}=\frac{\frac{9}{2}}{12}$$

Express \frac{\frac{9}{2}}{12} as a single fraction.

$$z^{2}=\frac{9}{2\times 12}$$

Multiply 2 and 12 to get 24.

$$z^{2}=\frac{9}{24}$$

Reduce the fraction \frac{9}{24} to lowest terms by extracting and canceling out 3.

$$z^{2}=\frac{3}{8}$$

Take the square root of both sides of the equation.

$$z=\frac{\sqrt{6}}{4} z=-\frac{\sqrt{6}}{4}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(1+2z\right)^{3}.

$$1+6z+12z^{2}+8z^{3}-\left(z+2\right)\left(z^{2}-2z+4\right)=3\left(\frac{3}{2}+2z\right)+7\left(z^{3}-1\right)$$

Use the distributive property to multiply z+2 by z^{2}-2z+4 and combine like terms.

$$1+6z+12z^{2}+8z^{3}-\left(z^{3}+8\right)=3\left(\frac{3}{2}+2z\right)+7\left(z^{3}-1\right)$$

To find the opposite of z^{3}+8, find the opposite of each term.

$$1+6z+12z^{2}+8z^{3}-z^{3}-8=3\left(\frac{3}{2}+2z\right)+7\left(z^{3}-1\right)$$

Combine 8z^{3} and -z^{3} to get 7z^{3}.

$$1+6z+12z^{2}+7z^{3}-8=3\left(\frac{3}{2}+2z\right)+7\left(z^{3}-1\right)$$

Subtract 8 from 1 to get -7.

$$-7+6z+12z^{2}+7z^{3}=3\left(\frac{3}{2}+2z\right)+7\left(z^{3}-1\right)$$

Use the distributive property to multiply 3 by \frac{3}{2}+2z.

$$-7+6z+12z^{2}+7z^{3}=\frac{9}{2}+6z+7\left(z^{3}-1\right)$$

Use the distributive property to multiply 7 by z^{3}-1.

$$-7+6z+12z^{2}+7z^{3}=\frac{9}{2}+6z+7z^{3}-7$$

Subtract 7 from \frac{9}{2} to get -\frac{5}{2}.

$$-7+6z+12z^{2}+7z^{3}=-\frac{5}{2}+6z+7z^{3}$$

Subtract -\frac{5}{2} from both sides.

$$-7+6z+12z^{2}+7z^{3}-\left(-\frac{5}{2}\right)=6z+7z^{3}$$

The opposite of -\frac{5}{2} is \frac{5}{2}.

$$-7+6z+12z^{2}+7z^{3}+\frac{5}{2}=6z+7z^{3}$$

Subtract 6z from both sides.

$$-7+6z+12z^{2}+7z^{3}+\frac{5}{2}-6z=7z^{3}$$

Add -7 and \frac{5}{2} to get -\frac{9}{2}.

$$-\frac{9}{2}+6z+12z^{2}+7z^{3}-6z=7z^{3}$$

Combine 6z and -6z to get 0.

$$-\frac{9}{2}+12z^{2}+7z^{3}=7z^{3}$$

Subtract 7z^{3} from both sides.

$$-\frac{9}{2}+12z^{2}+7z^{3}-7z^{3}=0$$

Combine 7z^{3} and -7z^{3} to get 0.

$$-\frac{9}{2}+12z^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$12z^{2}-\frac{9}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 0 for b, and -\frac{9}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{0±\sqrt{0^{2}-4\times 12\left(-\frac{9}{2}\right)}}{2\times 12}$$

Square 0.

$$z=\frac{0±\sqrt{-4\times 12\left(-\frac{9}{2}\right)}}{2\times 12}$$

Multiply -4 times 12.

$$z=\frac{0±\sqrt{-48\left(-\frac{9}{2}\right)}}{2\times 12}$$

Multiply -48 times -\frac{9}{2}.

$$z=\frac{0±\sqrt{216}}{2\times 12}$$

Take the square root of 216.

$$z=\frac{0±6\sqrt{6}}{2\times 12}$$

Multiply 2 times 12.

$$z=\frac{0±6\sqrt{6}}{24}$$

Now solve the equation z=\frac{0±6\sqrt{6}}{24} when ± is plus.

$$z=\frac{\sqrt{6}}{4}$$

Now solve the equation z=\frac{0±6\sqrt{6}}{24} when ± is minus.

$$z=-\frac{\sqrt{6}}{4}$$

The equation is now solved.

$$z=\frac{\sqrt{6}}{4} z=-\frac{\sqrt{6}}{4}$$