# Ex.2.3 Q5 Polynomials Solution - NCERT Maths Class 10

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## Question

Give examples of polynomials

$$p(x), \,g(x),\, q(x)$$ and $$r(x),$$ which satisfy the division algorithm and

(i)   deg $$p(x)=$$ deg $$q(x)$$

(ii)   deg $$q(x)=$$ deg $$r(x)$$

(iii)  deg $$r(x) =\,0$$

Video Solution
Polynomials
Ex 2.3 | Question 5

## Text Solution

What is known?

(i)   deg $$p(x)=$$ deg $$q(x)$$

(ii)   deg $$q(x)=$$ deg $$r(x)$$

(iii)  deg $$r(x) =\,0$$

What is unknown?

Examples of polynomials $$p(x),\, g(x), \,q(x)$$ and $$r(x),$$ which satisfy the division algorithm

Reasoning:

To solve this question, follow some steps

In case (i), assume polynomial $$p(x)$$ whose degree is equal to degree of $$q(x),$$ then put the values of $$p(x), \,g(x), \,q(x)$$ and $$r(x)$$

In the division algorithm, if L.H.S is equal to R.H.S, then the division algorithm is satisfied.

In case (ii), assume polynomial $$p(x)$$ in which degree of quotient $$q(x)$$ is equal to the degree of $$r(x),$$ then put the values of $$p(x), \,g(x),\, q(x)$$ and $$r(x)$$ in the division algorithm. If L.H.S is equal to R.H.S then the division algorithm is satisfied.

In case (iii), assume polynomial $$p(x)$$ in which degree of remainder $$r(x)$$ is equal to zero, then put the values of $$p(x),\, g(x),\, q(x)$$ and $$r(x)$$ in the division algorithm.

If L.H.S is equal to R.H.S, then the division algorithm is satisfied.

Use the below given statement of Division algorithm to solve this question

Division algorithm

Dividend $$=$$ Divisor $$\times$$ Quotient $$+$$ Remainder

According to division algorithm, if $$p(x)$$ and $$g(x)$$ are two polynomials with $$g\left( x \right) \ne 0,$$ then we can find polynomial $$q(x)$$ and $$r(x)$$ such that

$p\left( x \right)=g\left( x \right)\times q\left( x \right)+r\left( x \right)$

Where $$r\left( x \right) = 0$$ or degree of $$r(x) <$$ degree of $$g(x)$$

Degree of polynomial is the highest power of the variable in the polynomial.

Put the given values in the above equation and simplify it, get the value of $$g(x).$$

Steps:

(i)   deg $$p(x)=$$ deg $$q(x)$$

Degree of quotient will be equal to the degree of dividend when divisor is constant (i.e. when any polynomial is divided by a constant).

Let us assume the division of $$6{x^2} + {\text{ }}2x{\text{ }} + 2$$ by $$2$$

\begin{align} p\left( x \right)&=6{{x}^{2}}+2x\text{ }+2 \\ g\left( x \right)&=2 \\q\left( x \right)&=3{{x}^{2}}+\text{ }x+1,\; r\left( x \right)\text{ }=\text{ }0 \\ \end{align}

Degree of $$p(x)$$ and $$q(x)$$ is same i.e. $$2.$$

Checking for division algorithm:

\begin{align} p(x) &=g(x) \times q(x)+r(x) \\ 6 x^{2}+2 x+2 &=2\left(3 x^{2}+x+1\right)+0 \\ &=6 x^{2}+2 x+2 \end{align}

Thus, the division algorithm is satisfied.

(ii)   deg $$q(x)=$$ deg $$r(x)$$

Let us assume the division of $${x^3} + x$$ by $${x^2}$$

\begin{align}p\left( x \right)&={{x}^{3}}+\text{ }x \\ g\left( x \right)&={{x}^{2}} \\ q\left( x \right)\,&=x,\;r\left( x \right)\text{ }=x \\\end{align}

Clearly, degree of $$p(x)$$ and $$q(x)$$ is same i.e. $$1.$$

Checking for division algorithm

\begin{align}p\left( x \right)&=g\left( x \right)\times q\left( x \right)+r\left( x \right) \\{{x}^{3}}+x&=({{x}^{2}}\times x)+x \\{{x}^{3}}+x &= {{x}^{3}}+x \end{align}

Thus, the division algorithm is satisfied.

(iii)  deg $$r(x) =0$$

Degree of remainder will be 0 when remainder comes to a constant.

Let us assume the division of $${x^3} + 1$$ by $${x^2}.$$

$$p\left( x \right)={{x}^{3}}+1,\; \quad g\left( x \right)={{x}^{2}},\\ q\left( x \right)=\text{ }x \;\; \text{and} \;\; r\left( x \right)=1$$

Clearly, the degree of $$r(x)$$ is $$0.$$

Checking for division algorithm

\begin{align}p\left( x \right)&=g\left( x \right)\times q\left( x \right)+r\left( x \right) \\ {{x}^{3}}+1&=\text{ }({{x}^{2}}\times x)+1 \\ {{x}^{3}}+1&={{x}^{3}}+1 \\\end{align}

Thus, the division algorithm is satisfied.

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