Solve 7+2j^8+2j^3+29+2,9+3+3+3,1+3,1+3,2div10

$$7 + 2 j ^ { 8 } + 2 j ^ { 3 } + 29 + 2,9 + 3 + 3 + 3,1 + 3,1 + 3,2 \div 10$$

Evaluate

$$2j^{8}+2j^{3}+51,42$$

Solution Steps

Add 7 and 29 to get 36.

$$36+2j^{8}+2j^{3}+2,9+3+3+3,1+3,1+\frac{3,2}{10}$$

Add 36 and 2,9 to get 38,9.

$$38,9+2j^{8}+2j^{3}+3+3+3,1+3,1+\frac{3,2}{10}$$

Add 38,9 and 3 to get 41,9.

$$41,9+2j^{8}+2j^{3}+3+3,1+3,1+\frac{3,2}{10}$$

Add 41,9 and 3 to get 44,9.

$$44,9+2j^{8}+2j^{3}+3,1+3,1+\frac{3,2}{10}$$

Add 44,9 and 3,1 to get 48.

$$48+2j^{8}+2j^{3}+3,1+\frac{3,2}{10}$$

Add 48 and 3,1 to get 51,1.

$$51,1+2j^{8}+2j^{3}+\frac{3,2}{10}$$

Expand \frac{3,2}{10} by multiplying both numerator and the denominator by 10.

$$51,1+2j^{8}+2j^{3}+\frac{32}{100}$$

Reduce the fraction \frac{32}{100} to lowest terms by extracting and canceling out 4.

$$51,1+2j^{8}+2j^{3}+\frac{8}{25}$$

Add 51,1 and \frac{8}{25} to get \frac{2571}{50}.

$$\frac{2571}{50}+2j^{8}+2j^{3}$$

Factor

$$\frac{100j^{8}+100j^{3}+2571}{50}$$

Solution Steps

Add 7 and 29 to get 36.

$$factor(36+2j^{8}+2j^{3}+2,9+3+3+3,1+3,1+\frac{3,2}{10})$$

Add 36 and 2,9 to get 38,9.

$$factor(38,9+2j^{8}+2j^{3}+3+3+3,1+3,1+\frac{3,2}{10})$$

Add 38,9 and 3 to get 41,9.

$$factor(41,9+2j^{8}+2j^{3}+3+3,1+3,1+\frac{3,2}{10})$$

Add 41,9 and 3 to get 44,9.

$$factor(44,9+2j^{8}+2j^{3}+3,1+3,1+\frac{3,2}{10})$$

Add 44,9 and 3,1 to get 48.

$$factor(48+2j^{8}+2j^{3}+3,1+\frac{3,2}{10})$$

Add 48 and 3,1 to get 51,1.

$$factor(51,1+2j^{8}+2j^{3}+\frac{3,2}{10})$$

Expand \frac{3,2}{10} by multiplying both numerator and the denominator by 10.

$$factor(51,1+2j^{8}+2j^{3}+\frac{32}{100})$$

Reduce the fraction \frac{32}{100} to lowest terms by extracting and canceling out 4.

$$factor(51,1+2j^{8}+2j^{3}+\frac{8}{25})$$

Add 51,1 and \frac{8}{25} to get \frac{2571}{50}.

$$factor(\frac{2571}{50}+2j^{8}+2j^{3})$$

Factor out \frac{1}{50}. Polynomial 2571+100j^{8}+100j^{3} is not factored since it does not have any rational roots.

$$\frac{2571+100j^{8}+100j^{3}}{50}$$