Question

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

Answer 1

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