Find the zeros of the following quadratic polynomial: $15x^2 + 16x + 4$.

  • A
    $-1, -\frac{3}{5}$
  • B
    $-\frac{2}{3}, -\frac{2}{5}$
  • C
    $\frac{2}{3}, -\frac{1}{2}$
  • D
    $\frac{7}{6}, -\frac{1}{2}$

Explore More

Similar Questions

If $\sqrt{2}$ and $-\sqrt{2}$ are the zeros of $p(x) = 2x^{4} + 7x^{3} - 19x^{2} - 14x + 30$,then find the other zeros of $p(x)$.

Difficult
View Solution

Divide the following by the synthetic division method: $p(x) = x^{3} - 3x^{2} - 3x + 1$ by $x + 1$.

Are the following statements 'True' or 'False'? Justify your answers.
If the graph of a polynomial intersects the $x$-axis at only one point,it cannot be a quadratic polynomial.

Given that one of the zeroes of the cubic polynomial $ax^3 + bx^2 + cx + d$ is zero,the product of the other two zeroes is

State the degree of the given polynomial: $p(x) = x^{2} - \sqrt{3}x^{3} + 4x^{7} + 9$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo