Question

x²- 18x + 45. .............​

Answer 1

Given equation

$x^{2} -18x+45=0$

Solving, we get,

$x^{2} -18x+45=0$

$\implies x^{2} -3x-15x+45=0$

$\implies x(x-3)-15(x-3)=0$

$\implies (x-15)(x-3)=0$

Now,

$x+15=0$

$\implies x= 15$

$x-3=0$

$\implies x = 3$

$\boxed{\boxed{\bold{Therefore, \ the \ zeroes \ of \ the \ polynomial \ are \ 15 \ and \ 3}}}}}}$

                                                     

Answer 2

Given:-

• $\sf{ x^2 - 18x + 45}$

To find:-

• Zeroes of the polynomial.

Solution:-

$\bold{\underline{\boxed{\sf{\purple{ x^2 - 18x + 45}}}}}$

$\implies \sf{x^2 - 3x - 15x + 45 = 0}$

$\implies \sf{x(x - 3) -15(x - 3) = 0}$

$\implies \sf{(x - 15)(x - 3) = 0}$

$\implies \sf{either \; (x - 15) \; or \; (x - 3) \; = \; 0}$

$\bullet \sf{(x - 15) = 0}$

$\implies \sf{x = 15}$

$\bullet \sf{(x - 3) = 0}$

$\implies \sf{x = 3}$

$\bold{\underline{\underline{\boxed{\sf{\purple{Therefore, \; zeroes \; of \; the \; polynomial \; are \; 15 \; and \; 3}}}}}}$

$\rule{200}{2}$