MCQ
The number polynomials having zeroes as $-2$ and $5$ is :
- A$2$
- ✓More than $3$
- C$3$
- D$1$
✓
Answer
Correct option: B.
More than $3$
The number polynomials having zeroes as $-2$ and $5$ is more than $3$.
If $'S\ '$ is the sum and $'P\ ' $ is the product of the zeroes then the corresponding family of quadratic polynomial is given by $p(x)=k\left(x^2-S x+P\right)$ where $k$ is any real number.
$\therefore$ putting different values of $k,$ we can make more than $3$ numbers of polynomials.
If $'S\ '$ is the sum and $'P\ ' $ is the product of the zeroes then the corresponding family of quadratic polynomial is given by $p(x)=k\left(x^2-S x+P\right)$ where $k$ is any real number.
$\therefore$ putting different values of $k,$ we can make more than $3$ numbers of polynomials.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The number of polynomials having zeroes $-2$ and $5$ is:
→If the second term of an $AP$ is $13$ and its fifth term is $25,$ then its $7^{th}$ term is
→Find the roots of quadratic equation $6 x^2-13 x+5=0$
→in the given figure, two line segments $AC$ and $BD$ intersect each other at the point $P$ such that $PA = 6\ cm, PB = 3\ cm, PC = 2.5\ cm, PD = 5\ cm, \angle\text{APB}=50^\circ$ and $\angle\text{CDP}=30^\circ$ then $\angle\text{PBA}=?$
→If $\alpha,\ \beta$ be the zero of the polynomial $2x^2 + 5x + k$ such that $\alpha^2+\beta^2+\gamma^2=\frac{21}{4}$ then $k = ?$
→How many three $-$ digit numbers are divisible by $9$ ?
→The system $kx - y = 2$ and $6x - 2y = 3$ has a unique solution only when :
→Choose the correct answer from the given four options : A card is drawn from a deck of $52$ cards. The event $E$ is that card is not an ace of hearts. The number of outcomes favourable to $E$ is:
→$2$ One of the roots of the quadratic equation $ a^2 x^2-2 a b x+2 b^2=0$ is :
→The number of polynomials having zeros -3 and 5 is