1. clamin

|dw:1442446242052:dw|

2. anonymous

Almost, you forgot one thing: $$(-8i)(-3i)=24i^2$$

3. clamin

|dw:1442447294851:dw|

4. anonymous

Yes, unless $$i$$ is taken as the imaginary number, i.e. $$i=\sqrt{-1}$$. If that's the case, then $$i^2=-1$$, and so $$24i^2=-24$$.

5. clamin

|dw:1442447580763:dw| is this right??

6. clamin

that was my teacher's work

7. anonymous

Same problem here: the last term should be $$7i^2$$, or $$-7$$.

8. clamin

so the second problem is right?? right?

9. clamin

im just about the $i^{2}$

10. anonymous

Yes, the second problem is right, but it's missing that squared term. Include that and you're done. \begin{align*}4(6i)-(7i)(-5-i)&=24i+(7i)(5+i)\\ &=24i+35i+7i^2\\ &=59i+7i^2\\ &=59i-7 \end{align*}

11. clamin

what if you already have negative number and theres $i ^{2}$

12. clamin

for example $-10i ^{2}$

13. clamin

@sithsandgiggles

14. anonymous

$-10i^2=-10(-1)=10$