Question

Answer of the question X^2+5x-66

Answer 1

Given

$x^{2} +5x-66=0$

Solving, we get,

$x^{2} +5x-66=0$

$\implies x^{2} +11x-6x-66=0$

$\implies x(x+11)-6(x+11)=0$

$\implies (x-6)(x+11)=0$

Now,

$x-6=0\\\implies x = 6$

$x+11=0\\\implies x = -11$

$\boxed{\boxed{\bold{Therefore, \ 6 \ and \ -11 \ are \ the \ zeroes \ of \ the \ polynomial}}}}}}$

                                                             

Answer 2

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

x² + 5x - 66

$\bf{\red{\underline{\bf{Explanation\::}}}}$

Using by quadratic formula :

As the given polynomial compared with ax² + bx + c

• a = 1
• b = 5
• c = -66

Now;

$\longrightarrow\rm{x=\dfrac{-b\pm\sqrt{b^{2}-4ac } }{2a} }\\\\\\\longrightarrow\rm{x=\dfrac{-5\pm\sqrt{(5)^{2}-4\times 1\times (-66) } }{2\times 1} }\\\\\\\longrightarrow\rm{x=\dfrac{-5\pm\sqrt{25+264} }{2} }\\\\\\\longrightarrow\rm{x=\dfrac{-5\pm\sqrt{289} }{2} }\\\\\\\longrightarrow\rm{x=\dfrac{-5\pm17}{2} }\\\\\\\longrightarrow\rm{x=\dfrac{-5+17}{2} \:\:\:Or\:\:\:x=\dfrac{-5-17}{2} }\\\\\\\longrightarrow\m{x=\cancel{\dfrac{12}{2}} \:\:\:Or\:\:\:x=\cancel{\dfrac{-22}{2} }}$

$\longrightarrow\rm{\pink{x=6\:\:\Or\:\:x=-11}}$

Thus;

The zeroes of the polynomial will be x = 6 and x = -11 .