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ValentinkaMS [17]
3 years ago
13

Solve each equation. Check your solution. 7a+10=2a

Mathematics
1 answer:
Debora [2.8K]3 years ago
5 0

Answer:

a = -2

Step-by-step explanation:

7a + 10 = 2a

subtract 7a from each side

7a-7a+10 = 2a-7a

10 = -5a

divide by -5

10/-5 = -5a/-5

-2 = a

check

7 (-2) +10 = 2(-2)

-14 + 10 = -4

-4 = -4

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I can’t figure out how to use the zeros in the polynomial. Please explain
Temka [501]

Answer:

a) P (x) = (x + 3) (x-1) (x-4)

b) P (x) = (2x + 5) (5x - 4) (x-6)

c) P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2

Step-by-step explanation:

<u>For the question a *</u> you need to find a polynomial of degree 3 with zeros in -3, 1 and 4.

This means that the polynomial P(x) must be zero when x = -3, x = 1 and x = 4.

Then write the polynomial in factored form.

P (x) = (x + 3) (x-1) (x-4)

Note that this polynomial has degree 3 and is zero at x = -3, x = 1 and x = 4.

<u>For question b, do the same procedure</u>.

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The factors are

x = -\frac{5}{2}\\\\x +\frac{5}{2} = 0\\\\(2x +5) = 0

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x =\frac{4}{5}\\\\x-\frac{4}{5} = 0\\\\(5x-4) = 0

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

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

P (x) = (2x + 5) (5x - 4) (x-6)

<u>Finally for the question c we have</u>

Degree: 5

Zeros: -3, 1, 4, -1

Multiplicity 2 in -1

x = -3\\\\(x-3) = 0

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x = 1\\\\(x-1) = 0

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

x = 4\\\\(x-4) = 0

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x = -1\\\\(x + 1) = 0

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

P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2

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