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Tema [17]
3 years ago
8

Answer pleaseeeeeeeeee

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
5 0

Answer:

(g+5)²(g-3)

Step-by-step explanation:

we just need the denominator so we can ignore everything on top

with that in mind mulitply the denominators to find the LCD

(g²+2g-15)(g+5)=g³+7g²-5g-75

Factor this (i used synethic divison, can explain if needed)

(g+5)²(g-3)

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Answer:

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Step-by-step explanation:

2x³+ 6x² - x - 10 = 0

(1) Possible roots

The Rational Roots Theorem states that, if a polynomial has any rational roots, they will have the form p/q, where p is a factor of the constant term  and q is a factor of the leading coefficient.

\text{Possible rational root} = \dfrac{ p }{ q } = \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}

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\text{Possible root} = \dfrac{\text{factor of 10}}{\text{factor of 2}}

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\text{Possible roots are } \large \boxed{\mathbf{x = \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10}}

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Rather than work through all 12 possibilities, I will do one that works.

\begin{array}{r|rrrr}-2 & 2 & 6 & -1 & -10\\& & -4& -4 & 10\\& 2 & 2& -5 & 0\\\end{array}

So, x = -2 is a root, and the quotient is 2x² + 2x - 5.

(3) Check for other rational roots

2x² + 2x - 5 = 0

D = b² - 4ac =2²- 4(2)(-5) = 4 + 40 = 44

√44 = 2√11, which is irrational.

Since irrational roots come in pairs, the cubic equation has two real, irrational roots and one rational root at x = -2.

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