How do I solve the for x. 6x + 3 > 21

Mathematics

1 answer:

Iteru [2.4K]3 years ago

4 0

Answer:

x > 3

Step-by-step explanation:

6x + 3 > 21 (Move the constant, +3 to the right hand side and change its sign)

6x > 21 -3 (Subtract 21-3)

6x >18 (Divide both sides by 6)

x > 3

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Write a polynomial function with the given zeros. multiply the terms (write in standard form).

ryzh [129]

$1) \ f(x)=x^3-3x-2 \\ \\ 2) \ g(x)=x^3-x^2-6x$

<h2>Explanation:</h2>

$A \ \mathbf{polynomial \ function} \ of \ x \ with \ degree \ n \ is \ given \ by:\\ \\ f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots +a_{2}x^{2}+a_{1}x+a_{0} \\ \\ where \ n \ is \ a \ nonnegative \ integer \ and \ a_{n}, a_{n-1}, \ldots a_{2}, a_{1}, a_{0} \\ with \ a_{n}\neq 0$

In this exercise, we have two cases and both will be polynomial functions with degree 3 because we have three real roots in each case. So:

<h3>First.</h3>

The roots are:

$x= -1, -1, 2$

So we can write this polynomial functions as the product of linear factors:

$f(x)=(x-(-1))(x-(-1))(x-2) \\ \\ f(x)=(x+1)^2(x-2)$

Since we have to write it in standard form, let's expand:

$f(x)=(x+1)^2(x-2) \\ \\ \\ By \ property: \\ \\ (a+b)^2=a^2+2ab+b^2 \\ \\ \\ f(x)=(x^2+2x+1)(x-2) \\ \\ f(x)=x^3+2x^2+x-2x^2-4x-2 \\ \\ \\ Combine \ like \ terms: \\ \\ x^3+(2x^2-2x^2)+(x-4x)-2 \\ \\ \\ \boxed{f(x)=x^3-3x-2}$

<h3>Second.</h3>

The roots are:

$x= -2, 0, 3$

Writing the polynomial function as the product of linear factors:

$g(x)=(x-(-2))(x-0)(x-3) \\ \\ g(x)=(x+2)x(x-3) \\ \\ g(x)=x(x+2)(x-3) \\ \\ \\ Distributive \property: \\ \\ g(x)=x(x^2-3x+2x-6) \\ \\ g(x)=x(x^2-x-6) \\ \\ \\ Distributive \ property \ again: \\ \\ \boxed{g(x)=x^3-x^2-6x}$