Question

How do you factor `20b + 32b ^ { 2} - 250`?

Answer 1

`2(2b-5)(8b+25)`

Explanation:

Arrange the terms in descending powers of `b`

`32b^2+20b -250`

`=2(16b^2 +10b-125)" "larr` common factor of `2`

`2(2b-5)(8b+25)`

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Factorising the quadratic trinomial:

Find factors of `16 and 125` which SUBTRACT to give `10`

`10` is quite small in comparison to `16 and 125`.

This is a clue that you will be working with 'middle factors', not the full `16` and the full `125`
There is some trial and error, try any factors.

`" "16" "125`
`" "darr" "darr`
`" "2" "5" "rarr 8xx5 =40`
`" "8" "25" "rarr 2 xx25 = ul50`
`color(white)(xxxx....xxxxxxxxxxxx)10" "larr:.` correct factors

Now that you have the correct factors, consider the signs.
You must have `+10`

`" "2" "5" "rarr 8xx5 =color(red)(-)40`
`" "8" "25" "rarr 2 xx25 = color(red)(+)ul50`
`color(white)(xxxx..xxxxxxxxxxxx)color(red)(+)10`

Insert the signs into the working:

`" "2" "color(red)(-)5" "rarr 8xxcolor(red)(-)5 =color(red)(-)40`
`" "8" "color(red)(+)25" "rarr 2 xxcolor(red)(+)25 = color(red)(+)ul50`
`color(white)(xxxx..xxxxxxxxxxxxxxxx)color(red)(+)10`

Now you can use the rows to write down the factors:

`(2b-5)(8b+25)`

Answer 2

`20b+32b^2-250 = 2(2b-5)(8b+25)`

Explanation:

Given:

`20b+32b^2-250`

We can factor this by completing the square, then using the difference of squares identity:

`A^2-B^2 = (A-B)(A+B)`

with `A=(16b+5)` and `B=45`

First multiply by `8` to avoid some arithmetic involving fractions, then divide by `8` at the end...

`8(20b+32b^2-250) = 256b^2+160b-2000`

`color(white)(8(20b+32b^2-250)) = (16b)^2+2(16b)(5)+(5)^2-2025`

`color(white)(8(20b+32b^2-250)) = (16b+5)^2-45^2`

`color(white)(8(20b+32b^2-250)) = ((16b+5)-45)((16b+5)+45)`

`color(white)(8(20b+32b^2-250)) = (16b-40)(16b+50)`

`color(white)(8(20b+32b^2-250)) = 16(2b-5)(8b+25)`

Dividing both ends by `8`, we get:

`20b+32b^2-250 = 2(2b-5)(8b+25)`