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The solution to the given quadratic expression is x = -1 and x = 5
Factoring quadratic equation
Given the quadratic equation below
f(x) = x^2 - 4x - 5
Factorize
f(x) = x^2 - 5x + x - 5 = 0
f(x) = x(x - 5) + 1(x - 5) = 0
(x+1)(x-5) = 0
x = -1 and x = 5
Hence the solution to the given quadratic expression is x = -1 and x = 5
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Answer:
x = -1; x = 5
Step-by-step explanation:
Standard Form of a Quadratic Function: f(x) = ax² + bx + c, where a ≠ 0
Given function: f(x) = x² - 4x - 5
⇒ a = 1, b = -4, c = -5
We are asked to solve the following function by factoring. In order to do so, let's rewrite the middle term by finding the factors that give a product of the first and last terms (a • c = -5) and give us the sum of the middle term (b = -4).
Factors that give a product of a • c: 1 • -5 = -5
Factors that give a sum of b: 1 + (-5) = -4
Step 1: Substitute f(x) = 0.
⇒ 0 = x² - 4x - 5
Step 2: Rewrite the equation with the factors.
⇒ 0 = x² + x - 5x - 5
Step 3: Factor out x and -5.
⇒ 0 = (x² + x) + (-5x - 5)
⇒ 0 x(x + 1) - 5(x + 1) [ Factor out the common factor. ]
⇒ 0 = (x - 5)(x + 1)
Step 4: Apply the Zero-Product Property (if m•n = 0, then m = 0 or n = 0)
a) 0 = x - 5 ⇒ 0 + 5 = x - 5 + 5 ⇒ 5 = x
b) 0 = x + 1 ⇒ 0 - 1 = x + 1 - 1 ⇒ -1 = x
Therefore, the missing solution for this quadratic function is x = 5.
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