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3 years ago

According to the Fundamental Theorem of Algebra, how many roots exist for the polynomial function?

Mathematics

2 answers:

RideAnS [48]3 years ago

4 0

Three! -7/9, -1/4, and -4/3

garri49 [273]3 years ago

3 0

Answer with Step-by-step explanation:

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number can be considered a complex number with its imaginary part equal to zero.

Hence, the given polynomial equation has atleast 1 real root and atmost 3 real roots

(9x + 7)(4x + 1)(3x + 4) = 0

On solving this, we get

9x+7=0 or 4x+1=0 or 3x+4=0

i.e. x= -7/9 or x= -1/4 or x= -4/3

As we can see all the roots are real

Hence, number of real roots are:

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Somebody help me asap please giving brainliest!

Ierofanga [76]

Answer: 9/8 is greater than 7/10 because its greater than 1. 7/10 is not

Step-by-step explanation:^^^

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2 years ago

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Simplify each expression using the proper order of operations. please help thank you so much :)

Tasya [4]

Answer:

$\frac{41 - 3^2}{\sqrt{36} * 3 - 26} = -4$

$12 + 3\sqrt{8} * (9 - 2) = 12 + 42 \sqrt{2}$

$\frac{28 - (7^2 + 3)}{-13 + 3 * 5} = -12$

$7^2 - 5* 8+1 = 10$

$(2 * \sqrt{16}) - (\sqrt[3]{27} * \sqrt{81}) + 1 = -18$

Step-by-step explanation:

<em>Required: Solve the expressions using proper operation order</em>

To solve this, we'll make use of BODMAS

--------------------------------------------------------------------------------------------------------

$\frac{41 - 3^2}{\sqrt{36} * 3 - 26}$

Evaluate all squares and square roots

$\frac{41 - 3*3}{\sqrt{36} * 3 - 26}$

$\frac{41 - 3*3}{6 * 3 - 26}$

Evaluate the numerator (Start by multiplying 3 * 3)

$\frac{41 - 9}{6 * 3 - 26}$

Subtract 9 from 41

$\frac{32}{6 * 3 - 26}$

Evaluate the denominator (Start by multiplying 6 * 3)

$\frac{32}{18 - 26}$

$\frac{32}{-8}$

Divide 32 by -8

-4

Hence;

$\frac{41 - 3^2}{\sqrt{36} * 3 - 26} = -4$

--------------------------------------------------------------------------------------------------------

$12 + 3\sqrt{8} * (9 - 2)$

Start by evaluating the bracket

$12 + 3\sqrt{8} * 7$

Then evaluate the multiplication

$12 + 21\sqrt{8}$

Simplify the square root

$12 + 21\sqrt{4 * 2}$

Split the square root

$12 + 21\sqrt{4} * \sqrt{2}$

Take Square root of 4

$12 + 21 * 2} * \sqrt{2}$

$12 + 42 \sqrt{2}$

Hence;

$12 + 3\sqrt{8} * (9 - 2) = 12 + 42 \sqrt{2}$

--------------------------------------------------------------------------------------------------------

$\frac{28 - (7^2 + 3)}{-13 + 3 * 5}$

Evaluate 7²

$\frac{28 - (49 + 3)}{-13 + 3 * 5}$

Evaluate all expression in the bracket

$\frac{28 - (52)}{-13 + 3 * 5}$

$\frac{28 - 52}{-13 + 3 * 5}$

Evaluate 3 * 5

$\frac{28 - 52}{-13 + 1 5}$

$\frac{-2 4}{2}$

$-12$

Hence;

$\frac{28 - (7^2 + 3)}{-13 + 3 * 5} = -12$

--------------------------------------------------------------------------------------------------------

$7^2 - 5* 8+1$

Evaluate 7²

$49 - 5* 8+1$

Evaluate 5 * 8

$49 - 40 + 1$

$10$

$7^2 - 5* 8+1 = 10$

--------------------------------------------------------------------------------------------------------

$(2 * \sqrt{16}) - (\sqrt[3]{27} * \sqrt{81}) + 1$

Evaluate all square root and cube root

$(2 * 4) - (3 * 9) + 1$

Solve the expressions in bracket

$8 - 27 + 1$

$-18$

Hence;

$(2 * \sqrt{16}) - (\sqrt[3]{27} * \sqrt{81}) + 1 = -18$

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2 years ago

Rewrite log2^8=3 in exponential form

kolezko [41]

2^3=8 is log2^8=3 in exponential form

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3 years ago

What is 61 /<br> 100 -7/25

boyakko [2]

Answer:

0.33

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3 years ago

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2 2/3 divides by 1 3/5

Anarel [89]

1 2/3 . Hope this helps ;)

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3 years ago

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