How do you reduce $-8 + b^2$ by $5 + b^2$ ?

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Answer 1 · 2018-05-04T08:14:35

By subtracting $-13$

How do you reduce $20$ by $8$ ,

of course by $color(red)"subtracting "12$ , which means $20-8$ i.e. subtract latter number $8$ from former number $20$

Hence to reduce $-8+b^2$ by $5+b^2$ ,

we should subtract latter number $5+b^2$ from former number $-8+b^2$ i.e.

$-8+b^2-(5+b^2)$

= $-8+b^2-5-b^2$

= $-13$

i.e. by subtracting $-13$

How do you reduce -8 + b^2-8 + b^2 by 5 + b^25 + b^2?