Question

Show that $4$ and $1$ are the zeros of quadratic polynomial $p\left(x\right)=x^2-5x+4$. Also verify the relation between the zeros and the coefficient of the polynomial .

Answer 1

Comparing the given polynomial with $a{x}^2+bx+c$, we get

$a=1,b=5,c=4$

$\begin{array}{c}p\left(4\right)={\left(4\right)}^2-5\left(4\right)+4\\ =16+20+4\\ =0\\ p\left(1\right)={\left(1\right)}^2-5\left(1\right)+4\\ =1-5+4\\ =0\end{array}$

There ,$4,1$ are the zeroes of polynomial $x^2-5x+4$

So, we take $\alpha =4$ and $\beta =1$

Sum of zeroes:

$\begin{array}{c}\alpha +\beta =\frac{-b}{a}\\ \Rightarrow 4+1=\left(\frac{-5}{4}\right)\\ \Rightarrow 5=5\end{array}$

Product of zeroes:

$\begin{array}{c}\alpha \beta =\frac{c}{a}\\ \Rightarrow 4\times 1=\left(\frac{4}{1}\right)\\ \Rightarrow 4=4\end{array}$

Hence, proved