Solve =4m^4-29m^2n^2+25n^4

$$= 4 m ^ { 4 } - 29 m ^ { 2 } n ^ { 2 } + 25 n ^ { 4 }$$

Factor

$$\left(m-n\right)\left(2m-5n\right)\left(m+n\right)\left(2m+5n\right)$$

Solution Steps

Consider 4m^{4}-29m^{2}n^{2}+25n^{4} as a polynomial over variable m.

$$4m^{4}-29n^{2}m^{2}+25n^{4}$$

Find one factor of the form km^{p}+q, where km^{p} divides the monomial with the highest power 4m^{4} and q divides the constant factor 25n^{4}. One such factor is 4m^{2}-25n^{2}. Factor the polynomial by dividing it by this factor.

$$\left(4m^{2}-25n^{2}\right)\left(m^{2}-n^{2}\right)$$

Consider 4m^{2}-25n^{2}. Rewrite 4m^{2}-25n^{2} as \left(2m\right)^{2}-\left(5n\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2m-5n\right)\left(2m+5n\right)$$

Consider m^{2}-n^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(m-n\right)\left(m+n\right)$$

Rewrite the complete factored expression.

$$\left(m-n\right)\left(m+n\right)\left(2m-5n\right)\left(2m+5n\right)$$

Evaluate

$$4m^{4}+25n^{4}-29\left(mn\right)^{2}$$