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3 years ago

The variables a, b, and c represent polynomials where a = x 1, b = x2 2x − 1, and c = 2x. what is ab c in simplest form?

2 answers:

7 0

If you would like to write a * b + c in simplest form, you can do this using the following steps:

a = x + 1
b = x^2 + 2x - 1
c = 2x
a * b + c = (x + 1) * (x^2 + 2x - 1) + 2x = x^3 + 2x^2 - x + x^2 + 2x - 1 + 2x = x^3 + 3x^2 + 3x - 1

The correct result would be x^3 + 3x^2 + 3x - 1.

QveST [7]3 years ago

5 0

easyer way its c...........

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Answer:

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Step-by-step explanation:

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Since area of a rectangle = $base * height$

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3 years ago

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nikitadnepr [17]

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Step-by-step explanation:

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4 years ago

For what value of k is there one solution to the given quadratic equation (k+1)x²+4kx+2=0.

andriy [413]

Answer:

If $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

The signal of $\bigtriangleup$ determines how many real roots an equation has:

$\bigtriangleup > 0$ : Two real and different solutions

$\bigtriangleup = 0$ : One real solution

$\bigtriangleup < 0$ : No real solutions

In this problem, we have the following second order polynomial:

$(k+1)x^{2} + 4kx + 2 = 0$ .

This means that $a = k+1; b = 4k; c = 2$

It has one solution if

$\bigtriangleup = 0$

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$16k^{2} - 8k - 8 = 0$

We can simplify by 8

$2k^{2} - k - 1 = 0$

The solution is:

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...

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