Factor the following quadratic expression n^2+ 4n-12

Mathematics

1 answer:

nirvana33 [79]3 years ago

3 0

Answer:

(n + 6)(n - 2)

Step-by-step explanation:

Given

n² + 4n - 12

Consider the factors of the constant term (- 12) which sum to give the coefficient of the n- term (+ 4)

The factors are + 6 and - 2 , since

6 × - 2 = - 12 and 6 - 2 = + 4

Then

n² + 4n - 12 = (n + 6)(n - 2) ← in factored form

You might be interested in

Fill in the blank with the phrase that makes the proof statement true.

melamori03 [73]

Answer:

the two angels are bisect each other

5 0

3 years ago

Find the complex zeros of x^3+27. write f in factored form

aliina [53]

$x^3+27=0
\\x^3+3^3=0
\\(x+3)(x^2-3x+9)=0
\\x^2-3x+9=0
\\a=1,b=-3,c=9
\\
\\x_{1,2}= \frac{-b\pm \sqrt{b^2-4ac} }{2a} = \frac{-(-3)\pm \sqrt{(-3)^2-4\times1\times9} }{2\times1} =\frac{3\pm \sqrt{-27} }{2} =\frac{3\pm 3\sqrt{3}i }{2} 
\\
\\(x+3)(x-\frac{3+ 3\sqrt{3}i }{2})(x-\frac{3- 3\sqrt{3}i }{2})=0

$