Factor 3xy+21x+9 completely

Mathematics

1 answer:

inn [45]2 years ago

3 0

Answer:

<h2>Hence the second option is the correct answer</h2>

Step-by-step explanation:

In this problem we are required to factor out the given terms as stated in the expression, what is obtainable is that we look for terms(greatest terms) that are common to all the terms in the expression and factor it out.

here is a way to do it

given the following expression

$3xy+21x+9$

From the expression above we can see that 3 is common to all the terms of the expression so that we have

$3(xy+7x+3)$

hence the second option is the correct answer

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Step-by-step explanation:

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2 years ago

The complex zeros of the following polynomial function. Write f in factored form.

zavuch27 [327]

Answer: $\bold{x=-3\qquad x=\dfrac{3+3i\sqrt3}{2}\qquad x=\dfrac{3-3i\sqrt3}{2}}$

<u>Step-by-step explanation:</u>

The formula for factoring a cubic is:

a³ + b³ = (a + b)(a² - ab + b²)

f(x) = x³ + 3³

= (x + 3)(x² - 3x + 9)

To find the zeros, set each factor equal to zero and solve.

0 = x + 3 0 = x² - 3x + 9

-3 = x use quadratic formula

$x^2-3x+9\quad \rightarrow \quad a=1, b=-3, c=9\\\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-3)\pm\sqrt{(-3)^2-4(1)(9)}}{2(1)}\\\\\\x=\dfrac{3\pm\sqrt{9-36}}{2}\\\\\\x=\dfrac{3\pm\sqrt{-27}}{2}\\\\\\x=\dfrac{3\pm3i\sqrt{3}}{2}$