Solved on Sep 27, 2023

Rewrite $16y^3 - 3 = y - 48y^2$ in factored form. Find the solution set. $(y+3)(4y-1)(4y+1)=0$ A. The solution set is $-3, \frac{1}{4}, -\frac{1}{4}$ .

STEP 1

Assumptions1. The given equation is $16 y^{3}-3=y-48 y^{}$ . . The factored form of the equation is $(y+3)(4 y-1)(4 y+1)=0$ .
3. We are asked to find the solution set.

STEP 2

We start by setting each factor in the factored form of the equation equal to zero. This is because a product of factors equals zero if and only if at least one of the factors is zero.
$y+=0,4y-1=0,4y+1=0$

STEP 3

olve the equation $y+3=0$ for $y$ .
$y = -3$

STEP 4

olve the equation $4y-1=0$ for $y$ .
$y = \frac{1}{4}$

STEP 5

olve the equation $4y+1=0$ for $y$ .
$y = -\frac{1}{4}$

STEP 6

The solution set is the set of all solutions to the equation. Therefore, the solution set is $\{-3, \frac{1}{4}, -\frac{1}{4}\}$ .
The correct choice is A. The solution set is $\{-3, \frac{1}{4}, -\frac{1}{4}\}$ .

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Rewrite $16y^3 - 3 = y - 48y^2$ in factored form. Find the solution set. $(y+3)(4y-1)(4y+1)=0$ A. The solution set is $-3, \frac{1}{4}, -\frac{1}{4}$ .

STEP 1

Assumptions1. The given equation is $16 y^{3}-3=y-48 y^{}$ . . The factored form of the equation is $(y+3)(4 y-1)(4 y+1)=0$ .
3. We are asked to find the solution set.

STEP 2

We start by setting each factor in the factored form of the equation equal to zero. This is because a product of factors equals zero if and only if at least one of the factors is zero.
$y+=0,4y-1=0,4y+1=0$

STEP 3

olve the equation $y+3=0$ for $y$ .
$y = -3$

STEP 4

olve the equation $4y-1=0$ for $y$ .
$y = \frac{1}{4}$

STEP 5

olve the equation $4y+1=0$ for $y$ .
$y = -\frac{1}{4}$

STEP 6

The solution set is the set of all solutions to the equation. Therefore, the solution set is $\{-3, \frac{1}{4}, -\frac{1}{4}\}$ .
The correct choice is A. The solution set is $\{-3, \frac{1}{4}, -\frac{1}{4}\}$ .

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Rewrite 16y3−3=y−48y216y^3 - 3 = y - 48y^216y3−3=y−48y2 in factored form. Find the solution set. (y+3)(4y−1)(4y+1)=0(y+3)(4y-1)(4y+1)=0(y+3)(4y−1)(4y+1)=0 A. The solution set is −3,14,−14-3, \frac{1}{4}, -\frac{1}{4}−3,41​,−41​.

STEP 1

Assumptions1. The given equation is 16y3−3=y−48y16 y^{3}-3=y-48 y^{}16y3−3=y−48y. . The factored form of the equation is (y+3)(4y−1)(4y+1)=0(y+3)(4 y-1)(4 y+1)=0(y+3)(4y−1)(4y+1)=0.3. We are asked to find the solution set.

STEP 2

We start by setting each factor in the factored form of the equation equal to zero. This is because a product of factors equals zero if and only if at least one of the factors is zero.y+=0,4y−1=0,4y+1=0y+=0,4y-1=0,4y+1=0y+=0,4y−1=0,4y+1=0

STEP 3

olve the equation y+3=0y+3=0y+3=0 for yyy.y=−3y = -3y=−3

STEP 4

olve the equation 4y−1=04y-1=04y−1=0 for yyy.y=14y = \frac{1}{4}y=41​

STEP 5

olve the equation 4y+1=04y+1=04y+1=0 for yyy.y=−14y = -\frac{1}{4}y=−41​

STEP 6

The solution set is the set of all solutions to the equation. Therefore, the solution set is {−3,14,−14}\{-3, \frac{1}{4}, -\frac{1}{4}\}{−3,41​,−41​}.The correct choice is A. The solution set is {−3,14,−14}\{-3, \frac{1}{4}, -\frac{1}{4}\}{−3,41​,−41​}.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Rewrite $16y^3 - 3 = y - 48y^2$ in factored form. Find the solution set. $(y+3)(4y-1)(4y+1)=0$ A. The solution set is $-3, \frac{1}{4}, -\frac{1}{4}$ .

Assumptions1. The given equation is $16 y^{3}-3=y-48 y^{}$ . . The factored form of the equation is $(y+3)(4 y-1)(4 y+1)=0$ .
3. We are asked to find the solution set.

We start by setting each factor in the factored form of the equation equal to zero. This is because a product of factors equals zero if and only if at least one of the factors is zero.
$y+=0,4y-1=0,4y+1=0$

olve the equation $y+3=0$ for $y$ .
$y = -3$

olve the equation $4y-1=0$ for $y$ .
$y = \frac{1}{4}$

olve the equation $4y+1=0$ for $y$ .
$y = -\frac{1}{4}$

The solution set is the set of all solutions to the equation. Therefore, the solution set is $\{-3, \frac{1}{4}, -\frac{1}{4}\}$ .
The correct choice is A. The solution set is $\{-3, \frac{1}{4}, -\frac{1}{4}\}$ .