Solve left(12x+7right)^2(3x+2)(2x+1)=3

$${ \left(12x+7 \right) }^{ 2 } (3x+2)(2x+1)=3$$

Solve for x (complex solution)

$$x=-\frac{5}{6}\approx -0.833333333$$

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=-\frac{\sqrt{2}i}{6}-\frac{7}{12}\approx -0.583333333-0.23570226i$$

$$x=\frac{\sqrt{2}i}{6}-\frac{7}{12}\approx -0.583333333+0.23570226i$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(12x+7\right)^{2}.

$$\left(144x^{2}+168x+49\right)\left(3x+2\right)\left(2x+1\right)=3$$

Use the distributive property to multiply 144x^{2}+168x+49 by 3x+2 and combine like terms.

$$\left(432x^{3}+792x^{2}+483x+98\right)\left(2x+1\right)=3$$

Use the distributive property to multiply 432x^{3}+792x^{2}+483x+98 by 2x+1 and combine like terms.

$$864x^{4}+2016x^{3}+1758x^{2}+679x+98=3$$

Subtract 3 from both sides.

$$864x^{4}+2016x^{3}+1758x^{2}+679x+98-3=0$$

Subtract 3 from 98 to get 95.

$$864x^{4}+2016x^{3}+1758x^{2}+679x+95=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 95 and q divides the leading coefficient 864. List all candidates \frac{p}{q}.

$$±\frac{95}{864},±\frac{95}{432},±\frac{95}{288},±\frac{95}{216},±\frac{95}{144},±\frac{95}{108},±\frac{95}{96},±\frac{95}{72},±\frac{95}{54},±\frac{95}{48},±\frac{95}{36},±\frac{95}{32},±\frac{95}{27},±\frac{95}{24},±\frac{95}{18},±\frac{95}{16},±\frac{95}{12},±\frac{95}{9},±\frac{95}{8},±\frac{95}{6},±\frac{95}{4},±\frac{95}{3},±\frac{95}{2},±95,±\frac{19}{864},±\frac{19}{432},±\frac{19}{288},±\frac{19}{216},±\frac{19}{144},±\frac{19}{108},±\frac{19}{96},±\frac{19}{72},±\frac{19}{54},±\frac{19}{48},±\frac{19}{36},±\frac{19}{32},±\frac{19}{27},±\frac{19}{24},±\frac{19}{18},±\frac{19}{16},±\frac{19}{12},±\frac{19}{9},±\frac{19}{8},±\frac{19}{6},±\frac{19}{4},±\frac{19}{3},±\frac{19}{2},±19,±\frac{5}{864},±\frac{5}{432},±\frac{5}{288},±\frac{5}{216},±\frac{5}{144},±\frac{5}{108},±\frac{5}{96},±\frac{5}{72},±\frac{5}{54},±\frac{5}{48},±\frac{5}{36},±\frac{5}{32},±\frac{5}{27},±\frac{5}{24},±\frac{5}{18},±\frac{5}{16},±\frac{5}{12},±\frac{5}{9},±\frac{5}{8},±\frac{5}{6},±\frac{5}{4},±\frac{5}{3},±\frac{5}{2},±5,±\frac{1}{864},±\frac{1}{432},±\frac{1}{288},±\frac{1}{216},±\frac{1}{144},±\frac{1}{108},±\frac{1}{96},±\frac{1}{72},±\frac{1}{54},±\frac{1}{48},±\frac{1}{36},±\frac{1}{32},±\frac{1}{27},±\frac{1}{24},±\frac{1}{18},±\frac{1}{16},±\frac{1}{12},±\frac{1}{9},±\frac{1}{8},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-\frac{1}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 864x^{4}+2016x^{3}+1758x^{2}+679x+95 by 3\left(x+\frac{1}{3}\right)=3x+1 to get 288x^{3}+576x^{2}+394x+95. Solve the equation where the result equals to 0.

$$288x^{3}+576x^{2}+394x+95=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 95 and q divides the leading coefficient 288. List all candidates \frac{p}{q}.

$$±\frac{95}{288},±\frac{95}{144},±\frac{95}{96},±\frac{95}{72},±\frac{95}{48},±\frac{95}{36},±\frac{95}{32},±\frac{95}{24},±\frac{95}{18},±\frac{95}{16},±\frac{95}{12},±\frac{95}{9},±\frac{95}{8},±\frac{95}{6},±\frac{95}{4},±\frac{95}{3},±\frac{95}{2},±95,±\frac{19}{288},±\frac{19}{144},±\frac{19}{96},±\frac{19}{72},±\frac{19}{48},±\frac{19}{36},±\frac{19}{32},±\frac{19}{24},±\frac{19}{18},±\frac{19}{16},±\frac{19}{12},±\frac{19}{9},±\frac{19}{8},±\frac{19}{6},±\frac{19}{4},±\frac{19}{3},±\frac{19}{2},±19,±\frac{5}{288},±\frac{5}{144},±\frac{5}{96},±\frac{5}{72},±\frac{5}{48},±\frac{5}{36},±\frac{5}{32},±\frac{5}{24},±\frac{5}{18},±\frac{5}{16},±\frac{5}{12},±\frac{5}{9},±\frac{5}{8},±\frac{5}{6},±\frac{5}{4},±\frac{5}{3},±\frac{5}{2},±5,±\frac{1}{288},±\frac{1}{144},±\frac{1}{96},±\frac{1}{72},±\frac{1}{48},±\frac{1}{36},±\frac{1}{32},±\frac{1}{24},±\frac{1}{18},±\frac{1}{16},±\frac{1}{12},±\frac{1}{9},±\frac{1}{8},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-\frac{5}{6}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 288x^{3}+576x^{2}+394x+95 by 6\left(x+\frac{5}{6}\right)=6x+5 to get 48x^{2}+56x+19. Solve the equation where the result equals to 0.

$$48x^{2}+56x+19=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 48 for a, 56 for b, and 19 for c in the quadratic formula.

$$x=\frac{-56±\sqrt{56^{2}-4\times 48\times 19}}{2\times 48}$$

Do the calculations.

$$x=\frac{-56±\sqrt{-512}}{96}$$

Solve the equation 48x^{2}+56x+19=0 when ± is plus and when ± is minus.

$$x=-\frac{\sqrt{2}i}{6}-\frac{7}{12} x=\frac{\sqrt{2}i}{6}-\frac{7}{12}$$

List all found solutions.

$$x=-\frac{1}{3} x=-\frac{5}{6} x=-\frac{\sqrt{2}i}{6}-\frac{7}{12} x=\frac{\sqrt{2}i}{6}-\frac{7}{12}$$

Solve for x

$$x=-\frac{5}{6}\approx -0.833333333$$

$$x=-\frac{1}{3}\approx -0.333333333$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(12x+7\right)^{2}.

$$\left(144x^{2}+168x+49\right)\left(3x+2\right)\left(2x+1\right)=3$$

Use the distributive property to multiply 144x^{2}+168x+49 by 3x+2 and combine like terms.

$$\left(432x^{3}+792x^{2}+483x+98\right)\left(2x+1\right)=3$$

Use the distributive property to multiply 432x^{3}+792x^{2}+483x+98 by 2x+1 and combine like terms.

$$864x^{4}+2016x^{3}+1758x^{2}+679x+98=3$$

Subtract 3 from both sides.

$$864x^{4}+2016x^{3}+1758x^{2}+679x+98-3=0$$

Subtract 3 from 98 to get 95.

$$864x^{4}+2016x^{3}+1758x^{2}+679x+95=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 95 and q divides the leading coefficient 864. List all candidates \frac{p}{q}.

$$±\frac{95}{864},±\frac{95}{432},±\frac{95}{288},±\frac{95}{216},±\frac{95}{144},±\frac{95}{108},±\frac{95}{96},±\frac{95}{72},±\frac{95}{54},±\frac{95}{48},±\frac{95}{36},±\frac{95}{32},±\frac{95}{27},±\frac{95}{24},±\frac{95}{18},±\frac{95}{16},±\frac{95}{12},±\frac{95}{9},±\frac{95}{8},±\frac{95}{6},±\frac{95}{4},±\frac{95}{3},±\frac{95}{2},±95,±\frac{19}{864},±\frac{19}{432},±\frac{19}{288},±\frac{19}{216},±\frac{19}{144},±\frac{19}{108},±\frac{19}{96},±\frac{19}{72},±\frac{19}{54},±\frac{19}{48},±\frac{19}{36},±\frac{19}{32},±\frac{19}{27},±\frac{19}{24},±\frac{19}{18},±\frac{19}{16},±\frac{19}{12},±\frac{19}{9},±\frac{19}{8},±\frac{19}{6},±\frac{19}{4},±\frac{19}{3},±\frac{19}{2},±19,±\frac{5}{864},±\frac{5}{432},±\frac{5}{288},±\frac{5}{216},±\frac{5}{144},±\frac{5}{108},±\frac{5}{96},±\frac{5}{72},±\frac{5}{54},±\frac{5}{48},±\frac{5}{36},±\frac{5}{32},±\frac{5}{27},±\frac{5}{24},±\frac{5}{18},±\frac{5}{16},±\frac{5}{12},±\frac{5}{9},±\frac{5}{8},±\frac{5}{6},±\frac{5}{4},±\frac{5}{3},±\frac{5}{2},±5,±\frac{1}{864},±\frac{1}{432},±\frac{1}{288},±\frac{1}{216},±\frac{1}{144},±\frac{1}{108},±\frac{1}{96},±\frac{1}{72},±\frac{1}{54},±\frac{1}{48},±\frac{1}{36},±\frac{1}{32},±\frac{1}{27},±\frac{1}{24},±\frac{1}{18},±\frac{1}{16},±\frac{1}{12},±\frac{1}{9},±\frac{1}{8},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-\frac{1}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 864x^{4}+2016x^{3}+1758x^{2}+679x+95 by 3\left(x+\frac{1}{3}\right)=3x+1 to get 288x^{3}+576x^{2}+394x+95. Solve the equation where the result equals to 0.

$$288x^{3}+576x^{2}+394x+95=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 95 and q divides the leading coefficient 288. List all candidates \frac{p}{q}.

$$±\frac{95}{288},±\frac{95}{144},±\frac{95}{96},±\frac{95}{72},±\frac{95}{48},±\frac{95}{36},±\frac{95}{32},±\frac{95}{24},±\frac{95}{18},±\frac{95}{16},±\frac{95}{12},±\frac{95}{9},±\frac{95}{8},±\frac{95}{6},±\frac{95}{4},±\frac{95}{3},±\frac{95}{2},±95,±\frac{19}{288},±\frac{19}{144},±\frac{19}{96},±\frac{19}{72},±\frac{19}{48},±\frac{19}{36},±\frac{19}{32},±\frac{19}{24},±\frac{19}{18},±\frac{19}{16},±\frac{19}{12},±\frac{19}{9},±\frac{19}{8},±\frac{19}{6},±\frac{19}{4},±\frac{19}{3},±\frac{19}{2},±19,±\frac{5}{288},±\frac{5}{144},±\frac{5}{96},±\frac{5}{72},±\frac{5}{48},±\frac{5}{36},±\frac{5}{32},±\frac{5}{24},±\frac{5}{18},±\frac{5}{16},±\frac{5}{12},±\frac{5}{9},±\frac{5}{8},±\frac{5}{6},±\frac{5}{4},±\frac{5}{3},±\frac{5}{2},±5,±\frac{1}{288},±\frac{1}{144},±\frac{1}{96},±\frac{1}{72},±\frac{1}{48},±\frac{1}{36},±\frac{1}{32},±\frac{1}{24},±\frac{1}{18},±\frac{1}{16},±\frac{1}{12},±\frac{1}{9},±\frac{1}{8},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-\frac{5}{6}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 288x^{3}+576x^{2}+394x+95 by 6\left(x+\frac{5}{6}\right)=6x+5 to get 48x^{2}+56x+19. Solve the equation where the result equals to 0.

$$48x^{2}+56x+19=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 48 for a, 56 for b, and 19 for c in the quadratic formula.

$$x=\frac{-56±\sqrt{56^{2}-4\times 48\times 19}}{2\times 48}$$

Do the calculations.

$$x=\frac{-56±\sqrt{-512}}{96}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=-\frac{1}{3} x=-\frac{5}{6}$$