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#1**+2 **

\(\text{in (a), for example, you are told that }p(-5)=0\\ \text{thus }-5 \text{ is a zero and }x+5 \text{ is a factor}\\ \text{so divide }p(x) \text{ by }(x+5) \text{ using polynomial division}\\ \dfrac{p(x)}{x+5} = 6 x^2+x-1\\ \text{this should be easy enough to factor by eye}\\ 6x^2 + x -1 = (3x -1)(2x+1) \\ \text{and it's easy enough to read off the zeros in this form as }\\ x = \dfrac 1 3,~x = -\dfrac 1 2\\ \text{follow this procedure for (b), which I leave to you}\)

.Rom Feb 18, 2019

#2**+4 **

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

CPhill Feb 18, 2019