# Find the zeroes of the Polynomial and verify the relation between zeroes of the polynomial and its coefficients

x^{2}+4 x -192 Find the zeroes of the Polynomial and verify the relation between zeroes of the polynomial and its coefficients
x^{2}+4 x -192

Step By Step Solution :

Let’s assume

p(x)=x^{2}+4x-192

For zeroes of p(x)=0, we have to use factorization :

p(x)=x^{2}+16x-12x-192
\implies x \times(x+16)-12 \times(x+16)
\implies (x-12) \cdot(x+16)=p(x)=0

For p(x)=0,

Either

x-12=0\\ \therefore x=12

Or

x+16=0 \\ \therefore x=-16

Hence, the roots of the polynomial :

x_{1}=-16 \\\\ x_{2}=12

If

x_{1},x_{2}

are roots of the polynomial

ax^{2}+bx+c=0

then, the relation :

Sum \ of \ roots = x_{1}+x_{2} = \frac{-b}{a}=\frac {Coefficient\  of \ x}{Coeffficient \ of \ x^{2}}
Product \ of \ roots =x_{1} \cdot x_{2}= \frac{c}{a}=\frac {Constant\ Term}{Coeffficient \ of\  x^{2}}

If -16 and 12 are zeroes of

x^{2}+4x-192

then,

Sum \ of  zeros =x_{1}+x_{2}=-16+12=\frac{-4}{1}

After subtract in left hand side:

-4=-4


Left hand side and right hand side are equal.

Now,

Product of zeroes=

-16\times 12=\frac{-192}{1}

After multiply in left hand side:

-192=-192

Left hand side and right hand side are equal.

Solution ::The zeroes of polynomial

x^{2}+4x-192

are -16 and 12:

Now,

sum of zeroes =-16+12=-4= \frac{-4}{1}=\frac{-~Coefficient~ of ~x}{Coefficient ~of~ x^{2}}

Again,

Product of zeroes =-16\times 12=-192=\frac{-192}{1}=\frac{Constant ~term}{Coefficient ~of~ x^{2}}.