What are the solutions to the equation 2h^4+17h^3+21h^2=h+7 ?

The solutions to the equation 2h^4+17h^3+21h^2=h+7 are h=-1,h= 1/2 ,h=-7

Find the zeros of 2h^4+17h^3+21h^2=h+7

The zeros of 2h^4+17h^3+21h^2=h+7 are h=-1,h= 1/2 ,h=-7

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2 h^{4} + 17 h^{3} + 21 h^{2} = h + 7

h = - 1, h = \frac{1}{2}, h = - 7

Decimal Notation

h = - 1, h = 0.5, h = - 7

Solution steps

2 h^{4} + 17 h^{3} + 21 h^{2} = h + 7

Move

7

to the left side

2 h^{4} + 17 h^{3} + 21 h^{2} - 7 = h

Move

h

to the left side

2 h^{4} + 17 h^{3} + 21 h^{2} - 7 - h = 0

Write in the standard form

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

2 h^{4} + 17 h^{3} + 21 h^{2} - h - 7 = 0

Factor

2 h^{4} + 17 h^{3} + 21 h^{2} - h - 7 : (h + 1)^{2} (2 h - 1) (h + 7)

(h + 1)^{2} (2 h - 1) (h + 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

h + 1 = 0 or 2 h - 1 = 0 or h + 7 = 0

Solve

h + 1 = 0 : h = - 1

Solve

2 h - 1 = 0 : h = \frac{1}{2}

Solve

h + 7 = 0 : h = - 7

The solutions are

h = - 1, h = \frac{1}{2}, h = - 7

4 x^{4} - 4 x^{2} + 1 = 0

y^{2} (4 - y^{2}) = 0

0 = x^{3} + x^{2} - 12 x

5 x^{4} = 0

x^{3} - 11 x^{2} + 24 x = 0

What are the solutions to the equation 2h^4+17h^3+21h^2=h+7 ?
The solutions to the equation 2h^4+17h^3+21h^2=h+7 are h=-1,h= 1/2 ,h=-7
Find the zeros of 2h^4+17h^3+21h^2=h+7
The zeros of 2h^4+17h^3+21h^2=h+7 are h=-1,h= 1/2 ,h=-7