Question

FAN & MEDAL to who helps me solve:
2f(5f-2)-10(f^2-3f+6)=-8f(f+4)+4(2f^2-7f)

Answer 1

@Directrix

Answer 2

You have: \[2x(5x-2)-10(x^2-3x+6)=-8x(x+4)+4(2x^2-7x)\]
Now we write the linear polynomial on the left hand side in standard form: \[26x-60=-8x(x+4)+4(2x^2-7x)\] Now we do the same on the right side:
\[26x-60=-60x\]
Add 60x +60 to both sides: \[86x=60\]
Divide both sides with 86 \[x=30/43\]

Answer 3

I used x instead of "f", this can simply be changed.

Answer 4

where is 26 coming from @tommynaiter

Answer 5

\[2f(5f-2)-10(f^2-3f+6)=-8f(f+4)+4(2f^2-7f)\]\[10f^2-4f-10f^2-30f-16=-8f^2-32+8f^2-28\]\[-34f-16=60\]

Answer 6

Lets look at the left side. We have \[2x(5x-2)-10(x^2-3x+6)\]
First i want to remove the brackets (). I just multiply in:
\[10x^2 -4x-10(x^2-3x+6)=10x^2-4x-10x^2+30x-60=-4x+30x-60\]
\[=26x-60\]

Answer 7

this is what i keep getting

Answer 8

You need to remember that it is -10. So in order to remove the negative, we have to change + to - and - to +, in the parentheses.
\[-10(x^2-3x+6)=10(-x^2+3x-6)=-10x^2+30x-60\]

Answer 9

i think im overthinking it because i dont get what you're saying

Answer 10

Can you follow me here? \[2x(5x-2)=2x*5x-2x*2=10x^2-4x\]
\[-10(x^2-3x+6)=-10*x^2+(-10)*(-3x)+6*(-10)=-10x^2+30x-60\]

Answer 11

yeah, on the first line you distributed, but then idk what you did on the second line, i usually go through and distribute the whole line, then the next like i combine like terms and thats what im having trouble with, combining them and then everything after