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Sagot :
Let's solve the problem step-by-step.
### Part (A): Identifying Possible Rational Roots
Using the Rational Root Theorem, the possible rational roots of the polynomial [tex]\(8x^3 + 50x^2 - 41x + 7 = 0\)[/tex] are given by the factors of the constant term (7) divided by the factors of the leading coefficient (8).
1. Factors of the constant term (+7):
[tex]\[\pm 1, \pm 7\][/tex]
2. Factors of the leading coefficient (+8):
[tex]\[\pm 1, \pm 2, \pm 4, \pm 8\][/tex]
Therefore, the possible rational roots are all possible fractions formed by dividing factors of the constant term by factors of the leading coefficient. This results in:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}, \pm 7, \pm \frac{7}{2}, \pm \frac{7}{4}, \pm \frac{7}{8} \][/tex]
So, the answer to part (A) is:
[tex]\[ \boxed{\pm 1, \pm 7, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{1}{4}, \pm \frac{7}{4}, \pm \frac{1}{8}, \pm \frac{7}{8}} \][/tex]
### Part (B): Using Synthetic Division
We are given that one rational root of the equation is -7.
Using synthetic division to verify that -7 is indeed a root:
[tex]\[ \begin{array}{r|rrrr} -7 & 8 & 50 & -41 & 7 \\ & & -56 & 42 & -7 \\ \hline & 8 & -6 & 1 & 0 \end{array} \][/tex]
1. Write coefficients: [tex]\(8, 50, -41, 7\)[/tex].
2. Bring down the leading coefficient: [tex]\(8\)[/tex].
3. Multiply -7 by 8, and add to 50: [tex]\(50 + (-56) = -6\)[/tex].
4. Multiply -7 by -6, and add to -41: [tex]\(-41 + 42 = 1\)[/tex].
5. Multiply -7 by 1, and add to 7: [tex]\(7 + (-7) = 0\)[/tex].
Since we get a remainder of 0, [tex]\( -7 \)[/tex] is confirmed as a root.
### Part (C): Solving the Quotient Polynomial
After synthetic division, we obtained the quotient polynomial:
[tex]\[ 8x^2 - 6x + 1 = 0 \][/tex]
To solve this quadratic equation, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 8 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 8 \cdot 1}}{2 \cdot 8} = \frac{6 \pm \sqrt{36 - 32}}{16} = \frac{6 \pm \sqrt{4}}{16} \][/tex]
[tex]\[ x = \frac{6 \pm 2}{16} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{6 + 2}{16} = \frac{8}{16} = \frac{1}{2} \][/tex]
[tex]\[ x = \frac{6 - 2}{16} = \frac{4}{16} = \frac{1}{4} \][/tex]
Combining this with the root [tex]\( -7 \)[/tex] we found earlier, the solution set is:
[tex]\[ \boxed{-7, \frac{1}{2}, \frac{1}{4}} \][/tex]
### Part (A): Identifying Possible Rational Roots
Using the Rational Root Theorem, the possible rational roots of the polynomial [tex]\(8x^3 + 50x^2 - 41x + 7 = 0\)[/tex] are given by the factors of the constant term (7) divided by the factors of the leading coefficient (8).
1. Factors of the constant term (+7):
[tex]\[\pm 1, \pm 7\][/tex]
2. Factors of the leading coefficient (+8):
[tex]\[\pm 1, \pm 2, \pm 4, \pm 8\][/tex]
Therefore, the possible rational roots are all possible fractions formed by dividing factors of the constant term by factors of the leading coefficient. This results in:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}, \pm 7, \pm \frac{7}{2}, \pm \frac{7}{4}, \pm \frac{7}{8} \][/tex]
So, the answer to part (A) is:
[tex]\[ \boxed{\pm 1, \pm 7, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{1}{4}, \pm \frac{7}{4}, \pm \frac{1}{8}, \pm \frac{7}{8}} \][/tex]
### Part (B): Using Synthetic Division
We are given that one rational root of the equation is -7.
Using synthetic division to verify that -7 is indeed a root:
[tex]\[ \begin{array}{r|rrrr} -7 & 8 & 50 & -41 & 7 \\ & & -56 & 42 & -7 \\ \hline & 8 & -6 & 1 & 0 \end{array} \][/tex]
1. Write coefficients: [tex]\(8, 50, -41, 7\)[/tex].
2. Bring down the leading coefficient: [tex]\(8\)[/tex].
3. Multiply -7 by 8, and add to 50: [tex]\(50 + (-56) = -6\)[/tex].
4. Multiply -7 by -6, and add to -41: [tex]\(-41 + 42 = 1\)[/tex].
5. Multiply -7 by 1, and add to 7: [tex]\(7 + (-7) = 0\)[/tex].
Since we get a remainder of 0, [tex]\( -7 \)[/tex] is confirmed as a root.
### Part (C): Solving the Quotient Polynomial
After synthetic division, we obtained the quotient polynomial:
[tex]\[ 8x^2 - 6x + 1 = 0 \][/tex]
To solve this quadratic equation, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 8 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 8 \cdot 1}}{2 \cdot 8} = \frac{6 \pm \sqrt{36 - 32}}{16} = \frac{6 \pm \sqrt{4}}{16} \][/tex]
[tex]\[ x = \frac{6 \pm 2}{16} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{6 + 2}{16} = \frac{8}{16} = \frac{1}{2} \][/tex]
[tex]\[ x = \frac{6 - 2}{16} = \frac{4}{16} = \frac{1}{4} \][/tex]
Combining this with the root [tex]\( -7 \)[/tex] we found earlier, the solution set is:
[tex]\[ \boxed{-7, \frac{1}{2}, \frac{1}{4}} \][/tex]
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