Factorizing equation?

[raven2k7]

hi, can someone please explain to me in working how they ended up with the answer * 12*2k^2+12 . ive tried many times and just not getting that answer

1×(−3)×(k+1)+5×0×k+(−3×k+5)×1×(k+4)−k×(−3)×(−3×k+5)−(k+4)×0×1−(k+1)×1×5=−12×k^2+12

thank you

 

[Deleted member 4993]

hi, can someone please explain to me in working how they ended up with the answer * 12*2k^2+12 . ive tried many times and just not getting that answer

1×(−3)×(k+1)+5×0×k+(−3×k+5)×1×(k+4)−k×(−3)×(−3×k+5)−(k+4)×0×1−(k+1)×1×5=−12×k^2+12

thank you

Since you have tried many times, please share one of your attempts with us.

Please do not use 'x' as multiplication sign - use '*' instead.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

 

[raven2k7]

1*(−3)*(k+1)+5*0*k+(−3*k+5)*1*(k+4)−k*(−3)*(−3*k+5)−(k+4)*0*1−(k+1)*1*5=−12*k^2+12
-3k^2-3k+ 10 (what i got for the first part the second part i got 9k^2+k-5+5 ( maybe this is where i went wrong) dont know

 

[Deleted member 4993]

1*(−3)*(k+1)+5*0*k+(−3*k+5)*1*(k+4)−k*(−3)*(−3*k+5)−(k+4)*0*1−(k+1)*1*5=−12*k^2+12
-3k^2-3k+ 10 (what i got for the first part the second part i got 9k^2+k-5+5 ( maybe this is where i went wrong) dont know

You are saying:

-3k^2 - 3k + 10 (what i got for the first part)​

What is the first part?​

Similarly -

What is the second part?​

Please show your work completely, step-by-step.

 

[HallsofIvy]

1*(−3)*(k+1)+5*0*k+(−3*k+5)*1*(k+4)−k*(−3)*(−3*k+5)−(k+4)*0*1−(k+1)*1*5=−12*k^2+12
-3k^2-3k+ 10 (what i got for the first part the second part i got 9k^2+k-5+5 ( maybe this is where i went wrong) dont know

1*(-3)*(k+1)= -3*(k+1)= -3k- 3.
5*0*k= 0*k= 0
(-3*k+ 5)*1*(k+ 4)= (-3k+ 5)(k+ 4)= -3k(k+ 4)+ 5(k+ 4)= -3k^2- 12k+ 5k+ 20= -3k^2- 7k+ 20
k*(-3)(-3*k+5)= -3k(-3k+ 5)= (-3*-3)(k^2)- (-3k)5= 9k^2+ 15k.
-(k+4)*0*1= 0
-(k+1)*1*5= -5*(k+1)= -5k- 5

Add -3k- 3- 3k^2- 7k+ 20- 5k- 5= -3k^2- (3+7+ 5)k- (3- 20+ 5)= -3k^2- 15k+ 12

(And this problem does NOT involve "factorizing".)

 

[raven2k7]

1*(-3)*(k+1)= -3*(k+1)= -3k- 3.
5*0*k= 0*k= 0
(-3*k+ 5)*1*(k+ 4)= (-3k+ 5)(k+ 4)= -3k(k+ 4)+ 5(k+ 4)= -3k^2- 12k+ 5k+ 20= -3k^2- 7k+ 20
k*(-3)(-3*k+5)= -3k(-3k+ 5)= (-3*-3)(k^2)- (-3k)5= 9k^2+ 15k.
-(k+4)*0*1= 0
-(k+1)*1*5= -5*(k+1)= -5k- 5

Add -3k- 3- 3k^2- 7k+ 20- 5k- 5= -3k^2- (3+7+ 5)k- (3- 20+ 5)= -3k^2- 15k+ 12

(And this problem does NOT involve "factorizing".)

thank you sir very much appreciated .

 

[Deleted member 4993]

1*(-3)*(k+1)= -3*(k+1)= -3k- 3.
5*0*k= 0*k= 0
(-3*k+ 5)*1*(k+ 4)= (-3k+ 5)(k+ 4)= -3k(k+ 4)+ 5(k+ 4)= -3k^2- 12k+ 5k+ 20= -3k^2- 7k+ 20
k*(-3)(-3*k+5)= -3k(-3k+ 5)= (-3*-3)(k^2)- (-3k)5= 9k^2+ 15k.
-(k+4)*0*1= 0
-(k+1)*1*5= -5*(k+1)= -5k- 5

Add -3k- 3- 3k^2- 7k+ 20- 5k- 5= -3k^2- (3+7+ 5)k- (3- 20+ 5)= -3k^2- 15k+ 12

(And this problem does NOT involve "factorizing".)

Where did that go - in your final calculations?

According to your calculations then, \(\displaystyle L.H.S \ \ne \ R.H.S\) for OP?