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gregori [183]
2 years ago
5

Find the quotient of

%20%28x%20-%203%29%7D" id="TexFormula1" title="\frac{(x^{3} - 2x^{2} + 6x + 3)}{ (x - 3)}" alt="\frac{(x^{3} - 2x^{2} + 6x + 3)}{ (x - 3)}" align="absmiddle" class="latex-formula"> using long division
Mathematics
1 answer:
DedPeter [7]2 years ago
7 0
I can’t really do long division on a phone so, extended techniques are required.

(x^3 - 2x^2 + 6x + 3)/(x-3)

= (x^3 - 3x^2 + (x^2 - 6x + 9)+ 12x - 6)/(x-3)

= (x^2(x-3) + (x-3)^2 + 6(2x - 1))/x-3

= x^2 + x - 3 + 6(2x - 1)/(x-3)

or

= x^2 + x - 3 remainder 12x - 6


or, in the form p(x) = q(x)d(x) + r(x):


(x^3 - 2x^2 + 6x + 3) = (x^2 + x - 3)(x - 3) + (12x - 6)

which implies that the quotient q(x) = (x^2 + x - 3) and remainder r(x) = 12x - 6
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3 0
2 years ago
Solve 2x^2+x-4=0<br> X^2+_×+_=0
natima [27]

Answer:

Hopes it helps

Step-by-step explanation:

The Quadratic Polynomial is

2 x² +x -4=0

Using the Determinant method to find the roots of this equation

For, the Quadratic equation , ax²+ b x+c=0

(b) x²+x=0

x × (x+1)=0

x=0  ∧ x+1=0

x=0     ∧   x= -1

You can look the problem in other way

the two Quadratic polynomials are

2 x²+x-4=0, ∧ x²+x=0

x²= -x

So, 2 x²+x-4=0,

→ -2 x+x-4=0

→ -x -4=0

→x= -4

∨

x² +x² +x-4=0

x²+0-4=0→→x²+x=0

→x²=4

x=√4

x=2 ∧ x=-2

As, you will put these values into the equation, you will find that these values does not satisfy both the equations.

So, there is no solution.

You can solve these two equation graphically also.

3 0
2 years ago
URGENT HELP ME!! <br><br> please no links, thank you&lt;3
Illusion [34]

Answer:

x greater than or equal to 3

x greater than or equal to -8/3

Step-by-step explanation:

3 0
3 years ago
Write the number described by 1 ten 16 ones
Hatshy [7]
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3 years ago
1/2 ÷8 = <br><br> this on studyisland
vitfil [10]

Answer:

0.19

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
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