What is factor m^6+9m^4+24m^2+16 ?

The solution to factor m^6+9m^4+24m^2+16 is (m^2+1)(m^2+4)^2

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factor m^6+9m^4+24m^2+16

factor

m^{6} + 9 m^{4} + 24 m^{2} + 16

(m^{2} + 1) (m^{2} + 4)^{2}

Solution steps

m^{6} + 9 m^{4} + 24 m^{2} + 16

Let

u = m^{2}

= u^{3} + 9 u^{2} + 24 u + 16

Factor

u^{3} + 9 u^{2} + 24 u + 16 : (u + 1) (u + 4) (u + 4)

= (u + 1) (u + 4) (u + 4)

Substitute back

u = m^{2}

= (m^{2} + 1) (m^{2} + 4) (m^{2} + 4)

Apply exponent rule:

aa = a^{2}

(m^{2} + 4) (m^{2} + 4) = (m^{2} + 4)^{2}

= (m^{2} + 1) (m^{2} + 4)^{2}

2021 \cdot 2022 - 500 \cdot 4044 - 21 \cdot 2022

(3 y + x)^{3}

(- 2 x^{5})^{5} \cdot (- 5 x^{2})^{2}

x^{4} + 5 x^{3} - 5 x^{2} + 5 x - 6

16 x^{4} - 25 y^{6}

What is factor m^6+9m^4+24m^2+16 ?
The solution to factor m^6+9m^4+24m^2+16 is (m^2+1)(m^2+4)^2