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3 years ago

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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professor190 [17]

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Step-by-step explanation:

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3 years ago

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xenn [34]

Answer:

I think the first one

Step-by-step explanation:

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3 years ago

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liubo4ka [24]

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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

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

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

now, we can compare and find a , b and c

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

$x=-3+i$

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

$x=-3-i$

so, we will get solution

{−3+i, −3−i}.........Answer

(2)

we are given equation as

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

Since, Jamal solve this equation by completing square

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

$-8x=-2\times 4\times x$

so, we will add both sides by 4^2

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

we get

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