Trying to remember

[Robert Martin]

My son came to me for help with this math problem. He has the answer but I want to know how to arrive at it. It has been a while since I was in school so if someone could help me on how to work this problem it wolud be greatly appreciated.

15(2z+1)cubed+10(2z+1)squared-25(2z+1)

The answer he has is 20(z+1)(3z+4)

Please help

 

[soroban]

Hello, Robert Martin!

Factor: 15(2z + 1)³ + 10(2z + 1)² - 25(2z + 1)

The answer he has is: 20(z + 1)(3z + 4)

First, take out all common factors . . . each term has: 5(2z + 1)

. . . 5(2z + 1) [(3(2z + 1)² + 2(2z + 1) - 5]

That thing in the brackets can be factored, too, if we look at it right.

Let .u = 2z + 1, . then we have: . 3u² + 2u - 5

. . . which factors into: . (3u + 5)(u - 1)

Back-substitute: . (3[2z + 1] + 5) ([2z + 1] - 1)

. . . and we get: . (6z + 8)(2z) .= .2(3z + 4)(2z) .= .4z(3z + 4)


Hence, we have: . 5(2z + 1)·4z(3z + 4) . = . 20z(2z + 1)(3z + 4)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Isn't it much easier to read when I use spaces?

Of course, if I'm the <u>only</u> one using them, I'll stop.

I see nothing wrong with writing: x^3-8=(x-2)(x^2+4x+2) . . . do you?