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3 years ago

I need help with problem number 3 if someone can help me with that

Mathematics

1 answer:

Flura [38]3 years ago

3 0

Plug in x into the equation. Take x= -2 for example. Your equation is 2x-5, so instead of x, you put -2.
2(-2)-5 = -9
Your table is as follows:
(-2,-9)
(-1,-7)
(0,-5)
(1,-3)

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3 years ago

Givea)Possible number of positive real rootsb)Possible number of negative real rootsc)Possible rational roolsd)Find the roots

CaHeK987 [17]

The function is given to be:

$x^3-2x^2-3x+6$

QUESTION A

We can use Descartes' Rule of Signs to check the positive real roots of a polynomial.

The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number.

If we have:

$f(x)=x^3-2x^2-3x+6$

The coefficients are: +1, -2, -3, +6.

We can see that there are only 2 sign changes; from the first to the second term, and from the third to the fourth term.

Therefore, there are 2 or 0 positive real roots.

QUESTION B

To find the number of negative real roots, evaluate f(-x) and check for sign changes:

$\begin{gathered} f(-x)=(-x)^3-2(-x)^2-3(-x)+6 \\ f(-x)=-x-2x^2+3x+6 \end{gathered}$

The coefficients are: -1, -2, +3, +6.

We can see that there is only one sign change; from the second term to the third term.

Therefore, there is 1 negative real root.

QUESTION C

To check the possible rational roots, we can use the Rational Root Theorem since all the coefficients are integers.

The Rational Root Theorem states that if the polynomial:

$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0$

has any rational roots, they must be in the form:

$\Rightarrow\pm\mleft\lbrace\frac{factors\text{ of }a_0}{factors\text{ of }a_n}\mright\rbrace$

From the polynomial, the trailing coefficient is 6:

$a_o=6$

Factors of 6:

$\Rightarrow\pm1,\pm2,\pm3,\pm6$

The leading coefficient is 1:

$a_n=1$

Factors of 1:

$\Rightarrow\pm1_{}$

Write in the form

$\Rightarrow\mleft\lbrace\frac{a_o}{a_n}\mright\rbrace$

Therefore,

$\Rightarrow\pm(\frac{1}{1}),\pm(\frac{2}{1}),\pm(\frac{3}{1}),\pm(\frac{6}{1})$

Therefore, the possible rational roots are:

$\Rightarrow\pm1,\pm2,\pm3,\pm6$

QUESTION D

We can use a graph to check the roots of the polynomial. The graph is shown below:

The roots of the polynomial refer to the points when the graph intersects the x-axis.

Therefore, the roots of the polynomial are:

$x=-1.732,x=1.732,x=2$

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