Find the zeroes of the quadratic polynomial $f( x)=x^2-3x-28$.
Given: A quadratic polynomial $f( x)=x^2-3x-28$.
To do: To find the zeroes of the given polynomial.
Solution:
Given polynomial is $f( x)=x^2-3x-28$
$=x^2-7x+4x-28$
$=x( x-7)+4( x-7)$
$=( x-7)( x+4)$
Therefore $x=4,\ 7$
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