What are the solutions to the equation ((x+1)(2x+3)(x+2))/6 =((x+1)^2(2x+2))/6 ?

The solutions to the equation ((x+1)(2x+3)(x+2))/6 =((x+1)^2(2x+2))/6 are x=-1,x=-4/3

Find the zeros of ((x+1)(2x+3)(x+2))/6 =((x+1)^2(2x+2))/6

The zeros of ((x+1)(2x+3)(x+2))/6 =((x+1)^2(2x+2))/6 are x=-1,x=-4/3

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\frac{( x + 1 ) ( 2 x + 3 ) ( x + 2 )}{6} = \frac{( x + 1 ) ^{2} ( 2 x + 2 )}{6}

x = - 1, x = - \frac{4}{3}

Decimal Notation

x = - 1, x = - 1.33333 \dots

Solution steps

\frac{( x + 1 ) ( 2 x + 3 ) ( x + 2 )}{6} = \frac{( x + 1 ) ^{2} ( 2 x + 2 )}{6}

Multiply both sides by

6

(x + 1) (2 x + 3) (x + 2) = (x + 1)^{2} (2 x + 2)

Expand

(x + 1) (2 x + 3) (x + 2) : 2 x^{3} + 9 x^{2} + 13 x + 6

Expand

(x + 1)^{2} (2 x + 2) : 2 x^{3} + 6 x^{2} + 6 x + 2

2 x^{3} + 9 x^{2} + 13 x + 6 = 2 x^{3} + 6 x^{2} + 6 x + 2

Subtract

2 x^{3} + 6 x^{2} + 6 x + 2

from both sides

2 x^{3} + 9 x^{2} + 13 x + 6 - (2 x^{3} + 6 x^{2} + 6 x + 2) = 2 x^{3} + 6 x^{2} + 6 x + 2 - (2 x^{3} + 6 x^{2} + 6 x + 2)

Simplify

3 x^{2} + 7 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 3 \cdot 4}{2 \cdot 3}

7^{2} - 4 \cdot 3 \cdot 4 = 1

x_{1, 2} = \frac{- 7 \pm 1}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 7 + 1}{2 \cdot 3}, x_{2} = \frac{- 7 - 1}{2 \cdot 3}

x = \frac{- 7 + 1}{2 \cdot 3} : - 1

x = \frac{- 7 - 1}{2 \cdot 3} : - \frac{4}{3}

The solutions to the quadratic equation are:

x = - 1, x = - \frac{4}{3}

\frac{2 a ^{2} + 10 a}{3 a ^{2} + 15 a}

\frac{28}{- 4}

- \frac{5}{2} + 2

3 x + 1 \geq 5

x = 2^{- 3}

What are the solutions to the equation ((x+1)(2x+3)(x+2))/6 =((x+1)^2(2x+2))/6 ?
The solutions to the equation ((x+1)(2x+3)(x+2))/6 =((x+1)^2(2x+2))/6 are x=-1,x=-4/3
Find the zeros of ((x+1)(2x+3)(x+2))/6 =((x+1)^2(2x+2))/6
The zeros of ((x+1)(2x+3)(x+2))/6 =((x+1)^2(2x+2))/6 are x=-1,x=-4/3