CBSE Class 10 Maths Chapter 2 Polynomials solutions Ex 2.3 Math Solution || NCERT Solutions for Class 10 Maths Chapter 2: Polynomials (English Medium)

By long division method we have,

Quotient \( =x-3 \)
Remainder \( =7 x-9 \)

By long division method we have,

Quotient \( =x^{2}+x-3 \)
Remainder \( =8 \)

By long division method we have,

Quotient \( =-x^{2}-2 \)
Remainder \( =-5 x+10 \)

\( t^{2}-3=t^{2}+0 t-3 \)

Since the remainder is 0 ,
Hence, \( t^{2}-3 \) is a factor of \( 2 t^{4}+3 t^{3}-2 t^{2}-9 t-12 \).

Since the remainder is 0 ,
Hence, \( x^{2}+3 x+1 \) is a factor of \( 3 x^{4}+5 x^{3}-7 x^{2}+2 x+2 \).

Since the remainder \( \neq 0 \),
Hence, \( x^{3}-3 x+1 \) is not a factor of \( x^{5}-4 x^{3}+x^{2}+3x+1 \).

\( p(x)=3 x^{4}+6 x^{3}-2 x^{2}-10 x-5 \)
Since the two zeroes are \( \sqrt{\frac{5}{3}} \) and \( -\sqrt{\frac{5}{3}} \)
\( \therefore\left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)=\left(x^{2}-\frac{5}{3}\right) \) is a factor of \( 3 x^{4}+6 x^{3}-2 x^{2}-10 x-5 \).
Therefore, we divide the given polynomial by \( x^{2}-\frac{5}{3} \).

We know, Dividend \( =( \)Divisor \( \times \) quotient\( )+ \) remainder
\(3 x^{4}+6 x^{3}-2 x^{2}-10 x-5=\left(x^{2}-\frac{5}{3}\right)\left(3 x^{2}+6 x+3\right)\)
\( 3 x^{4}+6 x^{3}-2 x^{2}-10 x-5=\left(x^{2}-\frac{5}{3}\right)\left(x^{2}+2 x+1\right) \)
As \( (a+b)^{2}=a^{2}+b^{2}+2 a b \)
So, \( x^{2}+2 x+1=(x+1)^{2} \)
\( 3 x^{4}+6 x^{3}-2 x^{2}-10 x-5=3\left(x^{2}-\frac{5}{3}\right)(x+1)^{2} \) Therefore, its zero is given by \( x+1=0 \).
\( \Rightarrow x=-1,-1 \)
Hence, the zeroes of the given polynomial are \( \sqrt{\frac{5}{3}} \) and \( -\sqrt{\frac{5}{3}} \). and \(-1 , -1\) .

Given,
Polynomial, \( \mathrm{p}(x)=x^{3}-3 x^{2}+x+2 \) (dividend)
Quotient \( =(x-2) \)
Remainder \( =(-2 x+4) \)
To find: divisor \( =\mathrm{g}(x) \) we know, Dividend \( = \) Divisor \( \times \) Quotient + Remainder
\(\Rightarrow x^{3}-3 x^{2}+x+2=g(x) \times(x-2)+(-2 x+4)\)
\(\Rightarrow x^{3}-3 x^{2}+x+2+2 x-4=g(x)(x-2)\)
\(\Rightarrow x^{3}-3 x^{2}+3 x-2=g(x)(x-2)\)
\( \mathrm{g}(x) \) is the quotient when we divide \( \left(x^{3}-3 x^{2}+3 x-2\right) \) by \( (x-2) \)

\( \therefore \mathrm{g}(x)=\left(x^{2}-x+1\right) \)

Degree of a polynomial is the highest power of the variable in the polynomial. For example if \( f(x)=x^{3}-2 x^{2}+1 \), then the degree of this polynomial will be 3.
By division Algorithm : \( \mathrm{p}(x)=\mathrm{g}(x) \ x \ \mathrm{q}(x)+\mathrm{r}
(x) \)
It means when \( P(x) \) is divided by \( g(x) \) then quotient is \( q(x) \) and remainder is \( r(x) \) We need to start with \( p(x)=q(x) \) This means that the degree of polynomial \( \mathrm{p}(x) \) and quotient \( \mathrm{q}(x) \) is
same. This can only happen if the degree of \( g(x)=0 \) i.e \( p(x) \) is divided by a constant. Let \( p(x)=x^{2}+1 \) and \( g(x)=2 \)
\(\frac{p(x)}{g(x)}=\frac{x^{2}+1}{2}\)
The,
\(p(x)=g(x) \times\left(\frac{x^{2}+1}{2}\right)\)
Clearly, Degree of \( \mathrm{p}(x)= \) Degree of \(
\mathrm{q}(x) \)
2. Checking for division algorithm,
\(p(x)=g(x) \times q(x)+r(x)\)
\(=6 x^{2}+2 x+2=3\left(3 x^{2}+x+1\right)\)
\(=6 x^{2}+2 x+2\)
Thus, the division algorithm is satisfied.

Degree of a polynomial is the highest power of the variable in the polynomial. For example if \( f(x)=x^{3}-2 x^{2}+1 \), then the degree of this polynomial will be 3.
Let us assume the division of \( x^{3}+x \) by \( x^{2} \),
Here,
\(\mathrm{p}(x)=x^{3}+x\)
\(\mathrm{g}(x)=x^{2}\)
\(\mathrm{q}(x)=x \text { and } \mathrm{r}(x)=x\)
Clearly, the degree of \( q(x) \) and \( r(x) \) is the same i.e.,
Checking for division algorithm,
\(\mathrm{p}(x)=\mathrm{g}(x) \times
\mathrm{q}(x)+\mathrm{r}(x) x^{3}+x\)
\(=\left(x^{2}\right) \times x+x \)
\(=x^{3}+x=x^{3}+x\)
Thus, the division algorithm is satisfied.

Degree of a polynomial is the highest power of the variable in the polynomial. For example if \( f(x)=x^{3}-2 x^{2}+1 \), then the degree of this polynomial will be 3.
Degree of the remainder will be 0 when the remainder comes to a constant.
Let us assume the division of \( x^{3}+1 \) by \( x^{2} \).
Here,
\(p(x)=x^{3}+1 g(x)=x^{2}\)
\(q(x)=x \text { and } r(x)=1\)
Clearly, the degree of \( r(x) \) is 0 . Checking for division algorithm,
\(p(x)=g(x) \times q(x)+r(x) x^{3}+1\)
\(=\left(x^{2}\right) \times x+1 \ x^{3}+1=x^{3}+1\)
Thus, the division algorithm is satisfied.