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3 years ago

Find the zeroes of the polynomial function f(x)=x^4-5x^3+11x^2-25x+30.

Mathematics

1 answer:

katen-ka-za [31]3 years ago

7 0

Step-by-step explanation:

x⁴ − 5x³ + 11x² − 25x + 30

x⁴ − 5x³ + 6x² + 5x² − 25x + 30

x² (x² − 5x + 6) + 5 (x² − 5x + 6)

(x² + 5) (x² − 5x + 6)

(x² + 5) (x − 2) (x −3)

The zeros are 2, 3, and ±√5i.

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Please help! Due tonight!

Kay [80]

Answer:

B. 6,8,10

Step-by-step explanation:

to find a right triangle the equation: a^2 + b^2 = c^2, so just plug each number in to see.

6^2 + 8^2 = !0^2

36 + 64 = 100

100 = 100

Personally i knew almost immediately because 3,4,5 are the most common right triangles example and 6,8,10 are just a x2 of them so yea.

3 0

3 years ago

YO HELLO I NEED HELP WITH THIS

Tasya [4]

Answer- I don’t know sorry

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2 years ago

The graph of the function g(x)= x^2+3x-4 is shifted 5 units to the left

omeli [17]

<h3>THE MID TERM SPLITTING OF THE EQUATION IS :</h3>

${x}^{2} + 3x - 4$

${x}^{2} + 4x - x - 4$

$x(x + 4) - 1(x + 4)$

$x = -4 \: and \: 1$

$x = 1$

now shifted to 5 units left

therefore,

x=-4

hope it helps you mate thanks for the question and if possible please mark it as brainliest

5 0

3 years ago

What is the product?

Sophie [7]

Answer:

Step-by-step explanation:

In exponent multiplication, if base are same just add the powers.

$(-6a^{3}b+2ab^{2})(5a^{2}-2ab^{2}-b)=(-6a^{3}b)*(5a^{2}-2ab^{2}-b)+2ab^{2}*(5a^{2}-2ab^{2}-b)\\\\$

$=(-6a^{3}b)*5a^{2}-(-6a^{3}b)*2ab^{2}-(-6a^{3}b)*b+ 2ab^{2}*5a^{2}-2ab^{2}*2ab^{2}-2ab^{2}*b\\\\=-30a^{3+2}b+12a^{3+1}b^{1+2}+6a^{3}b^{1+1}+10a^{1+2}b^{2}-4a^{1+1}b^{2+2}-2ab^{2+1}\\\\=-30a^{5}b+12a^{4}b^{3}+6a^{3}b^{2}+10a^{3}b^{2}-4a^{2}b^{4}-2ab^{3}\\$