Expand
$(x+2 t)(x-5 t)$
Expand
$(x+2 t)(x-5 t)$
$(x+2 t)(x-5 t)$
$=(x)^{2}+(2 t-5 t) x+(2 t)(-5 t)$
$=x^{2}-3 x t-10 t^{2}$
Similar Questions
Dividing $x^{3}+125$ by $(x-5),$ the remainder is $\ldots \ldots \ldots .$
Dividing $x^{3}+125$ by $(x-5),$ the remainder is $\ldots \ldots \ldots .$
Verify whether the following are True or False:
$-3$ is a zero of $y^{2}+y-6.$
Verify whether the following are True or False:
$-3$ is a zero of $y^{2}+y-6.$
Check whether $g(x)$ is a factor of $p(x)$ or not, where
$p(x)=8 x^{3}-6 x^{2}-4 x+3, \quad g(x)=\frac{x}{3}-\frac{1}{4}$
Check whether $g(x)$ is a factor of $p(x)$ or not, where
$p(x)=8 x^{3}-6 x^{2}-4 x+3, \quad g(x)=\frac{x}{3}-\frac{1}{4}$
Find the value of each of the following polynomials at the indicated value of variables
$q(y)=5 y^{3}-4 y^{2}+14 y-\sqrt{3}$ at $y=2$
Find the value of each of the following polynomials at the indicated value of variables
$q(y)=5 y^{3}-4 y^{2}+14 y-\sqrt{3}$ at $y=2$
Without actually calculating the cubes, find the value of :
$\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$
Without actually calculating the cubes, find the value of :
$\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$