Answer:
[tex]\frac{3x^2+7x-20}{x+4}=3x-5[/tex]
Step-by-step explanation:
Long division:
Step 1:
We divide '3x²' by 'x' to get the first term of quotient.
⇒ [tex]\frac{3x^2}{x}=3x[/tex]
Now, we multiply '3x' to [tex](x+4)[/tex]
⇒ [tex]3x(x+4)=3x^2+12x[/tex]
Now, we subtract [tex]3x^2+12x[/tex] from the given numerator [tex]3x^2+7x-20[/tex]. This gives,
[tex]=3x^2+7x-20-(3x^2+12x)\\=3x^2+7x-20-3x^2-12x\\=3x^2-3x^2+7x-12x-20\\=-5x-20[/tex]
Step 2:
We divide the first term of the above result by 'x' again to get the second term of the quotient.
⇒ [tex]\frac{-5x}{x}=-5[/tex]
Now, we multiply '-5' to [tex](x+4)[/tex]
⇒ [tex]-5(x+4)=-5x-20[/tex]
Now, we subtract [tex]-5x-20[/tex] from the result obtained at the last of step 1 [tex]-5x-20[/tex]. This gives,
[tex]-5x-20-(-5x-20)=-5x+5x-20+20=0[/tex]
So, we stop division as we got a constant after subtraction. The constant is called the remainder and here the remainder is 0.
The quotient is our answer: [tex]3x-5[/tex]
