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Ronch [10]
3 years ago
7

What are the zeros of the polynomial function

Mathematics
1 answer:
solniwko [45]3 years ago
3 0

Option B

The zeros of the polynomial function are x = 0 , x = 6, x = -1

<em><u>Solution:</u></em>

Given function is:

f(x) = x^3 -5x^2 - 6x

<em><u>To find: zeros of the polynomial function</u></em>

To find the zeros of polynomial function, set the function equal to zero and then solve for x.

x^3 -5x^2 - 6x = 0

Taking "x" as common term, we get

x(x^2 - 5x - 6) = 0

Equating to zero,

x = 0 \text{ and } x^2 - 5x - 6 = 0

So one of the zeros of polynomial is x = 0

Let us solve x^2 - 5x - 6 = 0 to find other zeros

<em><u>Let us solve using quadratic formula,</u></em>

\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Using the above formula,

\text{ for } x^2 - 5x - 6 = 0 \text{ we have } a = 1 ; b = -5 ; c = -6

\text{ The discriminant } b^2 - 4ac > 0 , \text{ so, there are two real roots }

Substituting the values of a, b, c in above formula,

x=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-6)}}{2(1)}\\\\x=\frac{5 \pm \sqrt{25+24}}{2}=\frac{5 \pm \sqrt{49}}{2}\\\\x=\frac{5 \pm 7}{2}\\\\x=\frac{5+7}{2} \text{ or } x=\frac{5-7}{2}\\\\x=\frac{12}{2} \text{ or } x=\frac{-2}{2}\\\\x=6 \text{ or } x=-1

Thus the zeros of the polynomial function are x = 0 , x = 6, x = -1

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8 0
3 years ago
Rewrite the system of linear equations as a matrix equation AX = B.
iren2701 [21]

Answer:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

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Answer:

(0,12)

Step-by-step explanation:

Plug 0 into y

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