hi this is my question:
3x^2+4+x(x-1)
for this i know i would do
3x^2+4
but after that im kinda stuck!
would i do
3x^2+4 2x-x = 3x^2+x+4??
could you help please?
$\displaystyle x(x-1) = x \cdot x + x \cdot -1 = x^2-x$
Since multiplication is higher than addition in the order of operations you need to add $\displaystyle x^2-x$ such that you get $\displaystyle 3x^2+4+(x^2-x)$
Hi andyboy179,
you need to know how to multiply out the terms of x(x-1)
You do it the exact same way as 3(3-1)=3(2)=6
3(3-1)=3(3)+3(-1)=9-3=6
Or 5(5-1)=5(4)=20
5(5-1)=5(5)+5(-1)=25-5=20
You just multiply the x by both terms inside the brackets
$\displaystyle x(x-1)=x(x)+x(-1)=x^2-x$
Now combine that with $\displaystyle 3x^2+4$
Can you continue from there?