What is factor e^{9x}-5e^{3x}+4 ?

The solution to factor e^{9x}-5e^{3x}+4 is (e^x-1)(e^{2x}+e^x+1)(e^{6x}+e^{3x}-4)

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factor e^{9x}-5e^{3x}+4

factor

e^{9 x} - 5 e^{3 x} + 4

(e^{x} - 1) (e^{2 x} + e^{x} + 1) (e^{6 x} + e^{3 x} - 4)

Solution steps

e^{9 x} - 5 e^{3 x} + 4

Apply exponent rule:

a^{b c} = (a^{b})^{c}

e^{9 x} = (e^{3 x})^{3}

= (e^{3 x})^{3} - 5 e^{3 x} + 4

Use the rational root theorem

a_{0} = 4, a_{n} = 1

The dividers of

a_{0} : 1, 2, 4,

The dividers of

a_{n} : 1

Therefore, check the following rational numbers:

\pm \frac{1 , 2 , 4}{1}

\frac{1}{1}

is a root of the expression, so factor out

e^{3 x} - 1

= (e^{3 x} - 1) \frac{( e ^{3 x} ) ^{3} - 5 e ^{3 x} + 4}{e ^{3 x} - 1}

\frac{( e ^{3 x} ) ^{3} - 5 e ^{3 x} + 4}{e ^{3 x} - 1} = (e^{3 x})^{2} + e^{3 x} - 4

= (e^{3 x})^{2} + e^{3 x} - 4

Simplify

(e^{3 x} - 1) ((e^{3 x})^{2} + e^{3 x} - 4) : (e^{3 x} - 1) (e^{6 x} + e^{3 x} - 4)

= (e^{3 x} - 1) (e^{6 x} + e^{3 x} - 4)

Factor

e^{3 x} - 1 : (e^{x} - 1) (e^{2 x} + e^{x} + 1)

= (e^{x} - 1) (e^{x} + e^{2 x} + 1) (e^{6 x} + e^{3 x} - 4)

a + 3 b c + b + 3 a c

(x^{2}) + (4 x - 5)

a c + a d - b c - b d

n (A U B)

27 x^{3} y^{6} + z^{9}

What is factor e^{9x}-5e^{3x}+4 ?
The solution to factor e^{9x}-5e^{3x}+4 is (e^x-1)(e^{2x}+e^x+1)(e^{6x}+e^{3x}-4)