Solve the quadratic by factoring. x^(2)+4x-15=6x Answer: x=

Move Terms, Set Equal: First, we need to move all terms to one side of the equation to set it equal to zero. x^2 + 4x - 15 = 6x Subtract 6x from both sides to get: x^2 + 4x - 6x - 15 = 0 Combine like terms: x^2 - 2x - 15 = 0Factor Quadratic Equation: Now, we need to factor the quadratic equation x^2 - 2x - 15. We are looking for two numbers that multiply to -15 and add up to -2. The numbers -5 and 3 satisfy these conditions because: -5 * 3 = -15 -5 + 3 = -2 So we can write the factored form as: (x - 5)(x + 3) = 0Apply Zero Product Property: Next, we apply the zero product property, which states that if a product of two factors is zero, then at least one of the factors must be zero. So we set each factor equal to zero and solve for x: x - 5 = 0 or x + 3 = 0 Solving each equation gives us: x = 5 or x = -3

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Algebra 1

Factor quadratics with leading coefficient 1

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Q. Solve the quadratic by factoring.

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x^{2}+4 x-15=6 x

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Answer:

x=

Move Terms, Set Equal: First, we need to move all terms to one side of the equation to set it equal to zero. $\newline$ $x^2 + 4x - 15 = 6x$ $\newline$ Subtract $6x$ from both sides to get: $\newline$ $x^2 + 4x - 6x - 15 = 0$ $\newline$ Combine like terms: $\newline$ $x^2 - 2x - 15 = 0$
Factor Quadratic Equation: Now, we need to factor the quadratic equation $x^2 - 2x - 15$ . We are looking for two numbers that multiply to $-15$ and add up to $-2$ . The numbers $-5$ and $3$ satisfy these conditions because: $-5 \times 3 = -15$ $-5 + 3 = -2$ So we can write the factored form as: $(x - 5)(x + 3) = 0$
Apply Zero Product Property: Next, we apply the zero product property, which states that if a product of two factors is zero, then at least one of the factors must be zero. $\newline$ So we set each factor equal to zero and solve for $x$ : $\newline$ $x - 5 = 0$ or $x + 3 = 0$ $\newline$ Solving each equation gives us: $\newline$ $x = 5$ or $x = -3$