Answer:
(x+1)^2 = 49
x = 6 x = -8
Step-by-step explanation:
x^2 + 2x – 48 = 0
Add 48 to each side
x^2 + 2x =48
Take the coefficient of x
2
Divide by 2
2/2=1
Square it
1^2 =1
x^2 + 2x +1=48+1
x^2 +2x+1 = 49
(x+1)^2 = 49
Take the square root of each side
sqrt((x+1)^2) =±sqrt( 49)
x+1 = ±7
Subtract 1 from each side
x+1-1 = -1 ±7
x = -1+7 x = -1 -7
x = 6 x = -8
Answer:
1/2, 3
Step-by-step explanation:
This is a pretty involved problem, so I'm going to start by laying out two facts that our going to help us get there.
- The Fundamental Theorem of Algebra tells us that any polynomial has <em>as many zeroes as its degree</em>. Our function f(x) has a degree of 4, so we'll have 4 zeroes. Also,
- Complex zeroes come in pairs. Specifically, they come in <em>conjugate pairs</em>. If -2i is a zero, 2i must be a zero, too. The "why" is beyond the scope of this response, but this result is called the "complex conjugate root theorem".
In 2., I mentioned that both -2i and 2i must be zeroes of f(x). This means that both
and
are factors of f(x), and furthermore, their product,
, is <em>also</em> a factor. To see what's left after we factor out that product, we can use polynomial long division to find that

I'll go through to steps to factor that second expression below:

Solving both of the expressions when f(x) = 0 gets us our final two zeroes:


So, the remaining zeroes are 1/2 and 3.