What are the solutions to the equation 4(x+4)^4-53(x+4)^2=-49 ?

The solutions to the equation 4(x+4)^4-53(x+4)^2=-49 are x=-3,x=-5,x=-1/2 ,x=-15/2

Find the zeros of 4(x+4)^4-53(x+4)^2=-49

The zeros of 4(x+4)^4-53(x+4)^2=-49 are x=-3,x=-5,x=-1/2 ,x=-15/2

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4 (x + 4)^{4} - 53 (x + 4)^{2} = - 49

x = - 3, x = - 5, x = - \frac{1}{2}, x = - \frac{15}{2}

Decimal Notation

x = - 3, x = - 5, x = - 0.5, x = - 7.5

Solution steps

4 (x + 4)^{4} - 53 (x + 4)^{2} = - 49

Expand

4 (x + 4)^{4} - 53 (x + 4)^{2} : 4 x^{4} + 64 x^{3} + 331 x^{2} + 600 x + 176

4 x^{4} + 64 x^{3} + 331 x^{2} + 600 x + 176 = - 49

Move

49

to the left side

4 x^{4} + 64 x^{3} + 331 x^{2} + 600 x + 225 = 0

Factor

4 x^{4} + 64 x^{3} + 331 x^{2} + 600 x + 225 : (x + 3) (x + 5) (2 x + 1) (2 x + 15)

(x + 3) (x + 5) (2 x + 1) (2 x + 15) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 3 = 0 or x + 5 = 0 or 2 x + 1 = 0 or 2 x + 15 = 0

Solve

x + 3 = 0 : x = - 3

Solve

x + 5 = 0 : x = - 5

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

Solve

2 x + 15 = 0 : x = - \frac{15}{2}

The solutions are

x = - 3, x = - 5, x = - \frac{1}{2}, x = - \frac{15}{2}

0 = x^{3} - 2 x^{2} - 11 x + 12

a \cdot a \cdot a = 3 a

2 x^{3} + 3 x^{2} - 17 x + 12 = 0

z^{4} - 49 = 0

0.000004 x^{3} - 400 = 0

What are the solutions to the equation 4(x+4)^4-53(x+4)^2=-49 ?
The solutions to the equation 4(x+4)^4-53(x+4)^2=-49 are x=-3,x=-5,x=-1/2 ,x=-15/2
Find the zeros of 4(x+4)^4-53(x+4)^2=-49
The zeros of 4(x+4)^4-53(x+4)^2=-49 are x=-3,x=-5,x=-1/2 ,x=-15/2