Which point is an x-intercept of the quadratic function f(x) = (x – 4)(x + 2)?
1 answer:
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Answer:
1) There are four roots, with two real and two imaginary roots
The roots are x ±2, ±5·i
2) Therefore, there are 2 roots
The roots are real
The roots are x = -5 and x = 5
3) The possible factored form is therefore; (x + 5)²×(x + 1)×(x - 4)³×(x + 7)
4) Please see the attached graph of the function drawn with Microsoft Excel
Step-by-step explanation:
1) The given equations is as follows;
f(x) = x⁴ + 21·x² - 100
Let a = x², we have;
f(x) = a² + 21·a - 100
a = (-21 ± √(21² - 4 × 1 ×(-100))/(2 × 1)
a = -25 or 4
Therefore, x = √4 = ±2 or x = √(-25) = ±5·i
There are four roots, with two real and two imaginary roots
2) For f(x) = x³ - 5·x² - 25·x + 125
We have;
f(x) = x³ - 5·x² - 25·x + 125
From the equation, we see that x = 5 is a solution of the equation, therefore;
f(5) = 5³ - 5·5² - 25·5 + 125 = 0
Which gives, (x - 5) is a factor of the equation,
Dividing x³ - 5·x² - 25·x + 125 by (x - 5) gives;
x² - 25
(x³ - 5·x² - 25·x + 125)/(x - 5)
x³ - 5·x²
- 25·x
-25·x + 125
0 + 0
Therefore, by long division, (x³ - 5·x² - 25·x + 125)/(x - 5) = x² - 25
(x - 5)×(x² - 25) = (x - 5) × (x - 5) × (x + 5)
Therefore, there are 2 roots
The roots are real
The roots are x = -5 and x = 5
3) From the polynomial zeros, we have x = -5, x = -1, x = 4, and x = 7
At x = -5 the polynomial touches the x-axis given two real roots with (x + 5)² being a factor
With the root at x = -1, a factor is (x - 1)
With the root at x = 4 which has the shape of a cubic function, we have a factor of (x - 4)³
For the root at x = 1, the factor is taken as (x + 7)
The possible factored form is therefore; (x + 5)²×(x + 1)×(x - 4)³×(x + 7)
4) The given function is therefore;
f(x) = (x + 8)³(x + 6)²(x + 2)(x - 1)³(x - 3)⁴(x - 6)
(x + 8)³(x + 6)²(x + 2)(x - 1)³(x - 3)⁴(x - 6)
Please see the attached graph of the function drawn with Microsoft Excel
