Mn proof shows that ZDEH ZFEG in the attached diagram, but it is incomplete.
2 answers:
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Answer:
x = 5 and x = 1 are the factors
Step-by-step explanation:
<h2>step 1) </h2>identify the a, b, and c values (a=1 b= -6 c=5)
<h2>step 2) </h2>plug them into the quadratic formula
x = -(-6) ± √(-6)² - 4(1)(5) / 2(1)
<h2>step 3) </h2>multiply -1 by -6 to get 6
x = 6 ± √(-6)² - 4(1)(5) / 2(1)
<h2>step 4)</h2>multiply -6 by -6 to get 36
x = 6 ± √36 - 4(1)(5) / 2(1)
<h2>step 5)</h2>multiply 4 by 1 to get 4
x = 6 ± √36 - 4(5) / 2(1)
<h2>step 6) </h2>multiply 4 by 5 to get 20
x = 6 ± √36 - 20 / 2(1)
<h2>step 7)</h2>subtract 36 by 20 to get 16
x = 6 ± √16 / 2(1)
<h2>step 8) </h2>take the square root of 16 to get 4
x = 6 ± 4 / 2(1)
<h2>step 9)</h2>multiply 2 by 1 to get 2
x = 6 ± 4 / 2
<h2>step 10) </h2>add 6 by 4 to get 10
x = 10 / 2
<h2>step 11) </h2>divide 10 by 2 to get 5 (now you know the first factor is x=5)
x = 5
<h2>step 12)</h2>subtract 6 by 4 to get 2
x = 2 / 2
<h2>step 13)</h2>divide 2 by 2 to get 1 (now you have found the final factor x=1)
<h2>step 14)</h2>you know know that the factors of x²- 6x + 5 are<em><u> x = 5 and x = 1</u></em>
Answer:
Step-by-step explanation:
Given is an algebraic polynomial of degree 5.
Here leading term is p=3 and constant term is q =12
Factors of p are ±1,±2,±3
Factors of q are
Possible forms of p/q will be the same for any other polynomial of degree 5 with leading term =3 and constant term = 12
Hence any other polynomial
will have same possible zeroes of p/q, when a,b,c are rational.
Hence any polynomial of this type would have the same possible rational roots.