What is factor b^{4x}+b^{2x}-72 ?

The solution to factor b^{4x}+b^{2x}-72 is (b^{2x}-8)(b^{2x}+9)

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factor b^{4x}+b^{2x}-72

factor

b^{4 x} + b^{2 x} - 72

(b^{2 x} - 8) (b^{2 x} + 9)

Solution steps

b^{4 x} + b^{2 x} - 72

Apply exponent rule:

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

b^{4 x} = (b^{2 x})^{2}

= (b^{2 x})^{2} + b^{2 x} - 72

Break the expression into groups

= ((b^{2 x})^{2} - 8 b^{2 x}) + (9 b^{2 x} - 72)

Factor out

b^{2 x}

from

(b^{2 x})^{2} - 8 b^{2 x} : b^{2 x} (b^{2 x} - 8)

Factor out

9

from

9 b^{2 x} - 72 : 9 (b^{2 x} - 8)

= b^{2 x} (b^{2 x} - 8) + 9 (b^{2 x} - 8)

Factor out common term

b^{2 x} - 8

= (b^{2 x} - 8) (b^{2 x} + 9)

x^{2} - a x + b x - ab

144 - b^{8}

[- 1.5]

m p + m q + n p + n q

\frac{d}{d t} ((x) (x))

What is factor b^{4x}+b^{2x}-72 ?
The solution to factor b^{4x}+b^{2x}-72 is (b^{2x}-8)(b^{2x}+9)