(x%2B7).jpeg "(x-5)(x+7)")
Expanding and Solving (x-5)(x+7)
This article will explore how to expand and solve the expression (x-5)(x+7).
Expanding the Expression
To expand this expression, we can use the FOIL method, which stands for First, Outer, Inner, Last.
- First: Multiply the first terms of each binomial: x * x = x²
- Outer: Multiply the outer terms of the binomials: x * 7 = 7x
- Inner: Multiply the inner terms of the binomials: -5 * x = -5x
- Last: Multiply the last terms of each binomial: -5 * 7 = -35
Combining all the terms, we get: x² + 7x - 5x - 35
Simplifying the Expression
Next, we can simplify the expression by combining the like terms:
x² + 2x - 35
Solving the Expression
To solve the expression, we need to find the values of x that make the expression equal to zero. We can do this by factoring the expression or by using the quadratic formula.
Factoring:
- Find two numbers that multiply to -35 and add up to 2 (the coefficient of the x term). These numbers are 7 and -5.
- Rewrite the expression as (x + 7)(x - 5)
- Set each factor equal to zero and solve:
- x + 7 = 0 => x = -7
- x - 5 = 0 => x = 5
Therefore, the solutions to the equation (x-5)(x+7) = 0 are x = -7 and x = 5.
Quadratic Formula:
The quadratic formula can be used to solve any quadratic equation in the form ax² + bx + c = 0.
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Identify the coefficients: a = 1, b = 2, and c = -35.
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Substitute the values into the quadratic formula:
x = (-b ± √(b² - 4ac)) / 2a
x = (-2 ± √(2² - 4 * 1 * -35)) / 2 * 1
x = (-2 ± √(144)) / 2
x = (-2 ± 12) / 2
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Solve for x:
- x = (-2 + 12) / 2 = 5
- x = (-2 - 12) / 2 = -7
Therefore, the solutions to the equation (x-5)(x+7) = 0 are x = -7 and x = 5.
Conclusion
By expanding and solving the expression (x-5)(x+7), we can find the values of x that make the expression equal to zero. This can be achieved through factoring or using the quadratic formula.