Question Solve the equation $y^{3} + 3y^{2} = 4y + 12$ .

Studdy Solution

STEP 1

Assumptions
1. We are given the cubic equation $y^{3}+3y^{2}=4y+12$ .
2. We need to find all the real and/or complex solutions for the variable $y$ .

STEP 2

First, we want to set the equation to zero by moving all terms to one side of the equation.
$y^{3}+3y^{2}-4y-12=0$

STEP 3

Next, we look for any obvious factors of the cubic polynomial. We can try to use the Rational Root Theorem to find potential rational roots, which states that any rational root, expressed in its lowest terms $p/q$ , is such that $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient.

STEP 4

The constant term is -12, and the leading coefficient is 1. The factors of -12 are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$ . Since the leading coefficient is 1, any rational root must be a factor of the constant term.

STEP 5

We can use synthetic division or direct substitution to test these potential roots. Let's start by testing the smallest factors.

STEP 6

Test $y = 1$ :
$1^{3}+3(1)^{2}-4(1)-12 = 1+3-4-12 = -12 \neq 0$
So $y = 1$ is not a root.

STEP 7

Test $y = -1$ :
$(-1)^{3}+3(-1)^{2}-4(-1)-12 = -1+3+4-12 = -6 \neq 0$
So $y = -1$ is not a root.

STEP 8

Test $y = 2$ :
$2^{3}+3(2)^{2}-4(2)-12 = 8+12-8-12 = 0$
So $y = 2$ is a root.

STEP 9

Now that we have found a root, we can perform polynomial division or use synthetic division to factor $y - 2$ from the cubic polynomial.

STEP 10

Divide the polynomial $y^{3}+3y^{2}-4y-12$ by $y - 2$ .

STEP 11

Set up the division:
$\begin{array}{r|l} y-2 & y^{3}+3y^{2}-4y-12 \\ \end{array}$

STEP 12

Divide the first term of the polynomial $y^{3}$ by the first term of the divisor $y$ , which gives us $y^{2}$ . Multiply $y - 2$ by $y^{2}$ and subtract from the polynomial.
\begin{array}{r|ll} y-2 & y^{3} & +3y^{2} & -4y & -12 \\ & y^{3} & -2y^{2} & & \\ \cline{2-3} & & 5y^{2} & -4y & \\ \end{array}

STEP 13

Repeat the process: divide $5y^{2}$ by $y$ to get $5y$ , multiply $y - 2$ by $5y$ , and subtract from the remaining polynomial.
\begin{array}{r|lll} y-2 & y^{3} & +3y^{2} & -4y & -12 \\ & y^{3} & -2y^{2} & & \\ \cline{2-3} & & 5y^{2} & -4y & \\ & & 5y^{2} & -10y & \\ \cline{3-4} & & & 6y & -12 \\ \end{array}

STEP 14

Finally, divide $6y$ by $y$ to get $6$ , multiply $y - 2$ by $6$ , and subtract from the remaining polynomial.
\begin{array}{r|llll} y-2 & y^{3} & +3y^{2} & -4y & -12 \\ & y^{3} & -2y^{2} & & \\ \cline{2-3} & & 5y^{2} & -4y & \\ & & 5y^{2} & -10y & \\ \cline{3-4} & & & 6y & -12 \\ & & & 6y & -12 \\ \cline{4-5} & & & & 0 \\ \end{array}

STEP 15

The result of the division is $y^{2} + 5y + 6$ .

STEP 16

Now we have factored the original polynomial as:
$(y - 2)(y^{2} + 5y + 6) = 0$

STEP 17

Next, we need to factor the quadratic polynomial $y^{2} + 5y + 6$ .

STEP 18

We look for two numbers that multiply to $6$ and add up to $5$ . These numbers are $2$ and $3$ .

STEP 19

Factor the quadratic polynomial:
$y^{2} + 5y + 6 = (y + 2)(y + 3)$

STEP 20

Now we have the fully factored form of the original polynomial:
$(y - 2)(y + 2)(y + 3) = 0$

STEP 21

According to the Zero Product Property, if the product of several factors is zero, at least one of the factors must be zero.

STEP 22

Set each factor equal to zero and solve for $y$ :
$y - 2 = 0$ $y + 2 = 0$ $y + 3 = 0$

STEP 23

Solve each equation:
$y = 2$ $y = -2$ $y = -3$

STEP 24

We have found the three solutions to the original equation:
$y = 2, y = -2, y = -3$
These are the solutions to the equation $y^{3}+3y^{2}=4y+12$ .

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