# Q1 If -5 and 7 are zeroes of x^4 - 6x^3 - 26 x^2 + 138 x - 35 find the other zeroes.

1
by princey
Q2 If one zero of the polynomial 2x^2 - 3x +p is 3, then find the other root (zero).Also find the value of p.

2015-06-15T20:10:47+05:30
A1. Let f(x)= x⁴-6x³-26x²+138x-35
Zeros: 7 and -5
Therefore factors are: x-7 and x+5
Multiply x-7 and x+5 to get a combined factor (of bigger value/degree)
(x-7)(x+5)=x²-2x-35=g(x)
Divide f(x) by g(x)
_________________
x²-2x-35)x⁴-6x³-26x²+138x-35 (x²-4x+1
+x⁴-2x³-35x²
-    +    +
___________
-4x³+9x²+138x-35
-4x³+8x²+140x
+     -     -
_______________
x²-2x-35
+x²-2x-35
-    +   +
__________________
0

Now, quotient: q(x)=  x²-4x+1
Factorize it: (x-2-
√3)(x-2+√3)
=[x-(2+√3)][x-(2-√3)]
x=2+√3 or 2-√3
Therefore these are the other two zeros.

A2. f(x)= 2x²-3x+p
zero=3 So, factor= x-3=g(x)
Divide f(x) by g(x)
________
x-3)2x²-3x+p(2x+3=q(x)
+2x²-6x
-     +
_______
3x+p
+3x-9
-     +
_____
p+9=r(x)
So other factor: 2x+3
x=-3/2 is the other zero
but since g(x) is a factor of f(x)
The remainder r(x)=0
So, P+9=0
p=-9

Thank you...
M'pleasure :-)