Question

please answer me question number 14,15 both?

Answer 1

Answer:

Step-by-step explanation:

That means that the only possible rational zeros of f(x) are:

±12,±1,±32,±52,±3,±5,±152,±15

Trying the first one we find:

f(12)=28−54−282+15=1−5−56+604=0

So x=12 is a zero and (2x−1) a factor:

2x3−5x2−28x+15

=(2x−1)(x2−2x−15)

To factor the remaining quadratic, find a pair of factors of 15 which differ by 2. The pair 5,3works, so:

x2−2x−15=(x−5)(x+3)

So the two remaining zeros are x=5 and x=−3

Answer 2

14 no. and 15 no. answers
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