Find the zero(s) of the polynomial function f(x) = x2 - 36.

Mathematics

1 answer:

boyakko [2]1 year ago

3 0

Answer:

6 and -6

Explanations:

The given polynomial function is:

f(x) = x² - 36

The zeros of the polynomial are found by equation f(x) to zero

The equation becomes:

x² - 36 = 0

x² - 6² = 0

(x - 6) (x + 6) = 0

x - 6 = 0

x = 6

x + 6 = 0

x = -6

The zeros of the polynomial are -6 and 6

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Solve the equation: w^2+7w+12=0

polet [3.4K]

Hi there!
We are given the equation w² + 7w + 12 = 0, and we are told to solve it. Well, we can first take all the factors of 12 -
1 12
2 6
3 4
Now, take the sum of each factor pair -
1, 12 = 13
2, 6 = 8
3, 4 = 7
Find which factor pair adds up to 7, and we can see that 3 and 4 add up to seven, while also having a product of 12. Therefore, since the whole equation has addition signs, we can factor the equation w² + 7w + 12 into (w + 3)(w + 4) = 0. Next, using the Zero Product Property, we can set each term to zero.
w + 3 = 0
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w = -4
Therefore, the solution to the equation w² + 7w + 12 = 0 is w = -3, -4. Hope this helped and have a great day!

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Step-by-step explanation:

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3 years ago

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Solve for xxx. Enter the solutions from least to greatest. x^2 +7=43x 2 +7=43x, squared, plus, 7, equals, 43 \text{lesser }x =le

vaieri [72.5K]

Answer:

Step-by-step explanation:

x²+7=43x

x²-43x+7=0

$x=\frac{43 \pm \sqrt{(43)^2-4(1)(7)} }{2(1)} \\=\frac{43 \pm \sqrt{1849-28} }{2} \\=\frac{43 \pm \sqrt{1821} }{2} \\\approx \frac{43 \pm 42.673}{2}\\\approx 0.1635,42.8365$