Question

SOLVE USING LONG DIVISION:

Answer 1

Answer:

$\frac{3x^2+7x-20}{x+4}=3x-5$

Step-by-step explanation:

Long division:

Step 1:

We divide '3x²' by 'x' to get the first term of quotient.

⇒ $\frac{3x^2}{x}=3x$

Now, we multiply '3x' to $(x+4)$

⇒ $3x(x+4)=3x^2+12x$

Now, we subtract $3x^2+12x$ from the given numerator $3x^2+7x-20$. This gives,

$=3x^2+7x-20-(3x^2+12x)\\=3x^2+7x-20-3x^2-12x\\=3x^2-3x^2+7x-12x-20\\=-5x-20$

Step 2:

We divide the first term of the above result by 'x' again to get the second term of the quotient.

⇒ $\frac{-5x}{x}=-5$

Now, we multiply '-5' to $(x+4)$

⇒ $-5(x+4)=-5x-20$

Now, we subtract $-5x-20$ from the result obtained at the last of step 1 $-5x-20$. This gives,

$-5x-20-(-5x-20)=-5x+5x-20+20=0$

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: $3x-5$