Solve [4B^2+5BC]+[-2B^2-2BC-2C^2]+[2BC+4C^2]

$$[ 4 B ^ { 2 } + 5 B C ] + [ - 2 B ^ { 2 } - 2 B C - 2 C ^ { 2 } ] + [ 2 B C + 4 C ^ { 2 } ]$$

Evaluate

$$\left(B+2C\right)\left(2B+C\right)$$

Solution Steps

Combine 4B^{2} and -2B^{2} to get 2B^{2}.

$$2B^{2}+5BC-2BC-2C^{2}+2BC+4C^{2}$$

Combine 5BC and -2BC to get 3BC.

$$2B^{2}+3BC-2C^{2}+2BC+4C^{2}$$

Combine 3BC and 2BC to get 5BC.

$$2B^{2}+5BC-2C^{2}+4C^{2}$$

Combine -2C^{2} and 4C^{2} to get 2C^{2}.

$$2B^{2}+5BC+2C^{2}$$

Factor

$$\left(B+2C\right)\left(2B+C\right)$$

Solution Steps

Multiply and combine like terms.

$$2B^{2}+5BC+2C^{2}$$

Consider 2B^{2}+5BC+2C^{2} as a polynomial over variable B.

$$2B^{2}+5CB+2C^{2}$$

Find one factor of the form kB^{m}+n, where kB^{m} divides the monomial with the highest power 2B^{2} and n divides the constant factor 2C^{2}. One such factor is 2B+C. Factor the polynomial by dividing it by this factor.

$$\left(2B+C\right)\left(B+2C\right)$$