+6251 \(\text{in (a), for example, you are told that }p(-5)=0\\ \text{thus }-5 \text{ is a zero and }x+5 \text{ is a factor}\\ \text{so divide }p(x) \text{ by }(x+5) \text{ using polynomial division}\\ \dfrac{p(x)}{x+5} = 6 x^2+x-1\\ \text{this should be easy enough to factor by eye}\\ 6x^2 + x -1 = (3x -1)(2x+1) \\ \text{and it's easy enough to read off the zeros in this form as }\\ x = \dfrac 1 3,~x = -\dfrac 1 2\\ \text{follow this procedure for (b), which I leave to you}\)
.
+129852 Second one
We know that f(3) = 0.....so.....x = 3 is one zero
Using some synthetic division, we have
3 [ 2 1 -16 -15 ]
6 21 15
_________________________
2 7 5 0
The remaining polynomial is ( 2x^2 + 7x + 5)
To factor this......we ca break up 7x as 2x + 5x
2x^2 + 2x + 5x + 5 factor by grouping
2x ( x + 1) + 5 ( x + 1) =
(x + 1) (2x + 5)
Setting both factors to 0 and solving for x produces the other two zeros =
x = 1 and x = -5/2
