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