Question

# Obtain all zeroes of x4-3x3-7x2 +9x +12 iftwo of its zeroes are + V3.Sum of zeroes is

Solution

13 are zeroes = (x++3)(x - 3) = x2 – 3 is a divisor
of x4–3x3 - 7x2 +9x +12.17
x'-3x*-3x -7x +9x+12x-3x-4
4
x
-3x
3x³ 4x² +9x + 12
+9x
4x
4x
+ 12
+12
0
q(x) = x2 – 3x-4=x2-4x+x-4
=x(x-4)+1(x-4)
q(x) = (x+1)(x-4)
zeroes of q(x) are-1,4
.. Remaining required zeroes are -1,4.
Sum=-1+4=3