What are the solutions to the equation factor p^4+40=14p^2 ?

The solutions to the equation factor p^4+40=14p^2 are p=-2,p=2,p=-sqrt(10),p=sqrt(10)

Find the zeros of factor p^4+40=14p^2

The zeros of factor p^4+40=14p^2 are p=-2,p=2,p=-sqrt(10),p=sqrt(10)

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factor

p^{4} + 40 = 14 p^{2}

p = - 2, p = 2, p = - 10, p = 10

Decimal Notation

p = - 2, p = 2, p = - 3.16227 \dots, p = 3.16227 \dots

Solution steps

p^{4} + 40 = 14 p^{2}

Move

14 p^{2}

to the left side

p^{4} + 40 - 14 p^{2} = 0

Write in the standard form

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

p^{4} - 14 p^{2} + 40 = 0

Factor

p^{4} - 14 p^{2} + 40 : (p + 2) (p - 2) (p + 10) (p - 10)

(p + 2) (p - 2) (p + 10) (p - 10) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

p + 2 = 0 or p - 2 = 0 or p + 10 = 0 or p - 10 = 0

Solve

p + 2 = 0 : p = - 2

Solve

p - 2 = 0 : p = 2

Solve

p + 10 = 0 : p = - 10

Solve

p - 10 = 0 : p = 10

The solutions are

p = - 2, p = 2, p = - 10, p = 10

m^{4} + m^{3} + m^{2} = 0

x^{3} + 3 x^{2} - 24 x - 72 = 0

5 x^{3} + x^{2} - 20 x - 4 = 0

10 x^{4} - 20 x^{3} - 30 x^{2} = 0

6 x^{2} - 2 x^{3} = 0

What are the solutions to the equation factor p^4+40=14p^2 ?
The solutions to the equation factor p^4+40=14p^2 are p=-2,p=2,p=-sqrt(10),p=sqrt(10)
Find the zeros of factor p^4+40=14p^2
The zeros of factor p^4+40=14p^2 are p=-2,p=2,p=-sqrt(10),p=sqrt(10)