What are the solutions to the equation 6p^3+7p^2=96p+112 ?

The solutions to the equation 6p^3+7p^2=96p+112 are p=-7/6 ,p=-4,p=4

Find the zeros of 6p^3+7p^2=96p+112

The zeros of 6p^3+7p^2=96p+112 are p=-7/6 ,p=-4,p=4

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6 p^{3} + 7 p^{2} = 96 p + 112

p = - \frac{7}{6}, p = - 4, p = 4

Decimal Notation

p = - 1.16666 \dots, p = - 4, p = 4

Solution steps

6 p^{3} + 7 p^{2} = 96 p + 112

Move

112

to the left side

6 p^{3} + 7 p^{2} - 112 = 96 p

Move

96 p

to the left side

6 p^{3} + 7 p^{2} - 112 - 96 p = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

6 p^{3} + 7 p^{2} - 96 p - 112 = 0

Factor

6 p^{3} + 7 p^{2} - 96 p - 112 : (6 p + 7) (p + 4) (p - 4)

(6 p + 7) (p + 4) (p - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

6 p + 7 = 0 or p + 4 = 0 or p - 4 = 0

Solve

6 p + 7 = 0 : p = - \frac{7}{6}

Solve

p + 4 = 0 : p = - 4

Solve

p - 4 = 0 : p = 4

The solutions are

p = - \frac{7}{6}, p = - 4, p = 4

2 - 3 - 2 x^{4} - 3 x - 16 + 2 x - 3 = x + 12

- 12 x^{3} - 44 x^{2} - 40 x = 0

y^{3} = 3

4 x^{5} = 64 x^{3}

- r^{3} - 12 r^{2} + 60 r - 64 = 0

What are the solutions to the equation 6p^3+7p^2=96p+112 ?
The solutions to the equation 6p^3+7p^2=96p+112 are p=-7/6 ,p=-4,p=4
Find the zeros of 6p^3+7p^2=96p+112
The zeros of 6p^3+7p^2=96p+112 are p=-7/6 ,p=-4,p=4