All the real zeros of the given polynomial are integers. Find the zeros, and write...

interdicoxd

interdicoxd

Answered

2021-12-20

All the real zeros of the given polynomial are integers. Find the zeros, and write the polynomial in factored form.
P(x)=x36x2+12x8

Answer & Explanation

jgardner33v4

jgardner33v4

Expert

2021-12-21Added 35 answers

Step 1
Given:
P(x)=x36x2+12x8
Step 2
We can write the given polynomial as:
P(x)=(x)33(x)2(2)+3(x)(2)2(2)3
We have the formula,
(ab)3=a33a2b+3ab2b3
Hence,
P(x)=(x2)3
=(x-2)(x-2)(x-2)
This is the polynomial P(x) in factored form.
Step 3
To find the zeros of polynomial P(x), equate it to 0.
P(x)=0
(x-2)(x-2)(x-2)=0
x-2=0 or x-2=0 or x-2=0
x=2,2,2
Hence, the polynomial P(x) has all integer real zeros and they are equal to 2 with multiplicity 3.
Marcus Herman

Marcus Herman

Expert

2021-12-22Added 41 answers

The given polynomial is P(x)=x36x2+12x8
Since the leading coefficient is 1, any rational zero must be a divisior of the constant term -8.
So the possible rational zeros are ±1,±2,±4,±8
We test each of these possibilities
P(1)=(1)36(1)2+12(1)8
=1-6+12-8
=-1
P(1)=(1)36(1)2+12(1)8
=-1-6-12-8
=-27
P(2)=(2)3(6)(2)2+12(2)8
=8-24+24-8
=0
P(2)=(2)36(2)2+12(2)8
=-8-24-24-8
=-64
P(4)=(4)36(4)2+12(4)8
=64-96+48-8
=8
P(4)=(4)36(4)2+12(4)8
=-64-96-48-8
=-216
P(8)=(8)36(8)2+12(8)8
=512-384+96-8
=216
P(8)=(8)36(8)2+12(8)8
=-512-384-96-8
=-1000
The rational zero of P is 2

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