NCERT Solutions for Class 10 Maths Chapter 2: Polynomials || CBSE Class 10 Maths Chapter 2 Polynomials solutions Ex 2.4 Math Solution

\( P(x)=2 x^{3}+x^{2}-5 x+2 \)
Now for zeroes, putting the given values in \(x\).
\(\mathrm{P}(\frac{1 }{ 2})=2(\frac{1 }{ 2})^{3}+(\frac{1 }{ 2})^{2}-5(\frac{1 }{ 2})+2\)
\(=(\frac{1 }{ 4})+(\frac{1 }{ 4})-(\frac{5 }{ 2})+2=\frac{(1+1-10+8) }{ 2}=\frac{0 }{ 2}=0\)
\(P(1)=2 \times 1+1-5 \times 1+2=2+1-5+2=0\)
\(P(-2)=2 \times(-2)^{3}+(-2)^{2}-5(-2)+2\)
\(=(2 \times-8)+4+10+2=-16+16\)
\(=0\)
Thus, \( \frac{1 }{ 2},1 \) and \(-2\) are zeroes of given polynomial.
Comparing given polynomial with \( ax^{3}+bx^{2}+cx+\mathrm{d} \) and Taking zeroes as \( \alpha, \beta \), and \( \gamma \), we have
\( \mathrm{a}=2, \mathrm{~b}=1, \mathrm{c}=-5, \mathrm{~d}=2 \) and \( \alpha=\frac{1}{2}, \beta=1, \gamma=-2 \)
Now, We know the relation between zeroes and the coefficient of a standard cubic polynomial as
\(\alpha+\beta+\gamma=-\frac{b}{a}\)
Substituting value, we have
\(\frac{1}{2}+1-2=\frac{1}{2}\)
\(\frac{1}{2}=\frac{1}{2}\)
Since, LHS \( = \) RHS (Relation Verified)
\(\alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a}\)
\(\left(\frac{1}{2} \times 1\right)+(1 \times-2)+\left(-2 \times \frac{1}{2}\right)=-\frac{5}{2}\)
\(\frac{1}{2}-2-1=-\frac{5}{2}\)
\(-\frac{5}{2}=-\frac{5}{2}\)
Since LHS \( = \) RHS, Relation verified.
\(\alpha \beta \gamma=-\frac{d}{a}\)
\(\left(\frac{1}{2} \times 1 \times-2\right)=-\frac{2}{2}\)
\(-\frac{2}{2}=-\frac{2}{2}\)
Since LHS \( = \) RHS, Relation verified.
Thus, all three relationships between zeroes and the coefficient is verified.

\( p(x)=x^{3}-4 x^{2}+5 x-2 \)
Now for zeroes, put the given value in \( x \).
\(P(2)=2^{3}-4(2)^{2}+5 \times 2-2=8-16+10-2=18-18=0\)
\(P(1)=1^{3}-4(1)^{2}+5 \times 1-2=1-4+5-2=6-6=0\)
\(P(1)=1^{3}-4(1)^{2}+5 \times 1-2=1-4+5-2=6-6=0\)
Thus, 2, 1, 1 are the zeroes of the given polynomial.
Now,
Comparing the given polynomial with \( a x^{3}+b x^{2}+c x+d \), we get \( \mathrm{a}=1, \mathrm{~b}=-4 . \mathrm{c}=5, \mathrm{~d}=-2 \) and \( \alpha=2, \beta=1, \gamma=1 \)
Now,
\(
2+1+1=\frac{4}{1}\)
\(4=4\)
\(\alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a}\)
\((2 \times 1)+(1 \times 1)+(1 \times 2)=\frac{5}{1}\)
\(2+1+2=5\)
\(5=5\)
\(\alpha \beta \gamma=-\frac{d}{a}\)
\(2 \times 1 \times 1=2\)
\( 2=2 \)
Thus, all three relationships between zeroes and the coefficient is verified.

For a cubic polynomial equation, \( ax^{3}+bx^{2}+cx+\mathrm{d} \), and zeroes \( \alpha, \beta \) and \( \gamma \)
we know that
\(\alpha+\beta+\gamma=\frac{-b}{a}\)
\(\alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a}\)
\(\alpha \beta\gamma=\frac{-d}{a}\)
Let the polynomial be \( ax^{3}+bx+cx+\mathrm{d} \), and zeroes \( \alpha, \beta \) and \( \gamma \).
A cubic polynomial with respect to its zeroes is given by, \( x^{3}- \) (sum of zeroes) \( x^{2}+ \) (Sum of the product of roots taken two at a time) \( x- \) Product of Roots \( =0 \)
\(x^{3}-(2) x^{2}+(-7) x-(-14)=0\)
\(x^{3}-(2) x^{2}+(-7) x+14=0\)
Hence, the polynomial is \( x^{3}-2 x^{2}-7x+14 \).