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Kazeer [188]
3 years ago
6

Select all polynomials that have (x-3) as a factor.

Mathematics
1 answer:
Alchen [17]3 years ago
8 0

Answer:

Below.

Step-by-step explanation:

Substitute x = 3 into each polynomial and see if the value of the polynomial is zero. If it is zero then x-3 is a factor.

Try number 1:

a(x) = x^3 - 2x^2 - 4x + 3

a(3) = 3^3 - 2(3)^2 - 4(3) + 3

= 27 - 2(9) - 12 + 3

= 27 - 18 - 12 + 3

= 9 - 18 - 9

= 27 - 27

= 0

So a(x) has (x-3) is a factor of a(x)

The other 3 polynomials are solved in the same way.

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Answer:

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3 years ago
Please help quickly !
xenn [34]

Answer:

I think the first one

Step-by-step explanation:

because the first equation is written correctly

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liubo4ka [24]

3..3333333 this is the answer

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3 years ago
What is the solution set of the equation using the quadratic formula?
oee [108]

(1)

we are given

x^2+6x+10=0

we can use quadratic formula

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

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we get

a=1,b=6,c=10

now, we can plug these values into quadratic formula

x=\frac{-6\pm \sqrt{6^2-4(1)(10)}}{2(1)}

x=\frac{-6+\sqrt{6^2-4\cdot \:1\cdot \:10}}{2\cdot \:1}

we can simplify it

x=-3+i

x=\frac{-6-\sqrt{6^2-4\cdot \:1\cdot \:10}}{2\cdot \:1}

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{−3+i, −3−i}.........Answer

(2)

we are given equation as

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so, firstly we will move constant term on right side

so,  subtract both sides by 14

x^2-8x+14-14=0-14

x^2-8x=-14

we can write

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so, we will add both sides by 4^2

x^2-8x+4^2=-14+4^2

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x^2-8x+16=-14+16..............Answer


4 0
3 years ago
154=−4(8+6r) + 24r thanks to anyone who helps
vova2212 [387]

Answer:

When you do the math, you get an incorrect equation.

Step-by-step explanation:

154 = -32 -24r +24r

154 = -32

so you would get 186 = 0

4 0
2 years ago
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