Question

How do you divide $\left( {{x}^{3}}+3{{x}^{2}}-2x+5 \right)\div \left( x+1 \right)$ and identify any restriction on the variable ?

Answer 1

Hint: We can include any value to the range of the variable such that the denominator will be 0. So in this case x + 1 should not be equal to 0. When will divide $\left( {{x}^{3}}+3{{x}^{2}}-2x+5 \right)\div \left( x+1 \right)$ , the coefficient of x in the remainder should be less than coefficient of x in divisor.

Complete step-by-step answer:
Let’s divide $\left( {{x}^{3}}+3{{x}^{2}}-2x+5 \right)\div \left( x+1 \right)$ , if we look at first 2 term in dividend it is written ${{x}^{3}}+3{{x}^{2}}$ but ${{x}^{3}}+{{x}^{2}}$ is divisible by x + 1 so we can write $\left( {{x}^{3}}+3{{x}^{2}}-2x+5 \right)=\left( {{x}^{3}}+{{x}^{2}}+2{{x}^{2}}-2x+5 \right)$
Now if we look at second 2 terms in the dividend it written $2{{x}^{2}}-2x$ , but $2{{x}^{2}}+2x$ is divisible by x + 1 so we can write $\left( {{x}^{3}}+{{x}^{2}}+2{{x}^{2}}-2x+5 \right)=\left( {{x}^{3}}+{{x}^{2}}+2{{x}^{2}}+2x-4x+5 \right)$
Last 2 terms are -4x + 5 , -4x – 4 is divisible by x + 1
So we can write $\left( {{x}^{3}}+{{x}^{2}}+2{{x}^{2}}+2x-4x+5 \right)=\left( {{x}^{3}}+{{x}^{2}}+2{{x}^{2}}+2x-4x-4+9 \right)$
If we take ${{x}^{2}}$ common in first 2 terms and 2x common in second 2 terms and -4 common next 2 terms we will get
$\left( {{x}^{3}}+{{x}^{2}}+2{{x}^{2}}+2x-4x-4+9 \right)={{x}^{2}}\left( x+1 \right)+2x\left( x+1 \right)-4\left( x+1 \right)+9$
Now taking x + 1 common in first 3 terms we will get
$\left( {{x}^{3}}+{{x}^{2}}+2{{x}^{2}}+2x-4x-4+9 \right)=\left( x+1 \right)\left( {{x}^{2}}+2x-4 \right)+9$
So we can see that the remainder is 9 and quotient is $\left( {{x}^{2}}+2x-4 \right)$

Note: We can find the remainder by remainder theorem, when we divide f(x) by x – a , the reminder is f(a). We can check it by applying the formula to the given question. Hence the divisor is x + 1 the remainder will be f(-1) . so we will put – 1 in the equation $\left( {{x}^{3}}+3{{x}^{2}}-2x+5 \right)$. So the remainder is equal to $\left( {{\left( -1 \right)}^{3}}+3{{\left( -1 \right)}^{2}}-2\left( -1 \right)+5 \right)$ which is 9.