2x³+5x²-28x-15 zero are 3 plz solve full
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Explanation:
f(x)=2x3−5x2−28x+15
By the rational root theorem, since f(x) has integer coefficients, any rational zeros must be expressible in the form pq for integers p,q with pa divisor of the constant term 15 and q a divisor of the coefficient 2 of the leading term.
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
Explanation:
f(x)=2x3−5x2−28x+15
By the rational root theorem, since f(x) has integer coefficients, any rational zeros must be expressible in the form pq for integers p,q with pa divisor of the constant term 15 and q a divisor of the coefficient 2 of the leading term.
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
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