What is the answer to solvefor d,d(x)=(x^2+13x+42)/(x^2-3x-28) ?

The answer to solvefor d,d(x)=(x^2+13x+42)/(x^2-3x-28) is d=(x^2+13x+42)/(x(x^2-3x-28));\quad x\ne 0

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solvefor d,d(x)=(x^2+13x+42)/(x^2-3x-28)

solve for

d, d (x) = \frac{x ^{2} + 13 x + 42}{x ^{2} - 3 x - 28}

d = \frac{x ^{2} + 13 x + 42}{x ( x ^{2} - 3 x - 28 )}; x \neq = 0

Solution steps

d (x) = \frac{x ^{2} + 13 x + 42}{x ^{2} - 3 x - 28}

Divide both sides by

x; x \neq = 0

d = \frac{x ^{2} + 13 x + 42}{x ( x ^{2} - 3 x - 28 )}; x \neq = 0

y^{^{''}} - 7 y^{^{'}} = x^{2}

x^{^{''}} + 3 x^{^{'}} + 2 x = 3 cos (2 t)

2 x^{^{''}} + 5 x^{^{'}} + 4 x = 10 r (t)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = e^{x} - x^{3} + 16

x^{^{''}} + 6 x^{^{'}} - 4 x = 8 e^{t}

What is the answer to solvefor d,d(x)=(x^2+13x+42)/(x^2-3x-28) ?
The answer to solvefor d,d(x)=(x^2+13x+42)/(x^2-3x-28) is d=(x^2+13x+42)/(x(x^2-3x-28));\quad x\ne 0