Question

*1. $x(x - 7) = 18$

1. $x(2x + 37 ) = - 80$

1. $3x^5 = 20x^4 + 32x^3$

*4. $6x^3 + 3x^2 - 30x = 0$

Explanation:

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Problem 1: $x(x - 7) = 18$

Step1: Expand left-hand side

$x^2 - 7x = 18$

Step2: Rearrange to standard quadratic form

$x^2 - 7x - 18 = 0$

Step3: Factor the quadratic

$(x - 9)(x + 2) = 0$

Step4: Solve for $x$

Set each factor to 0: $x - 9 = 0$ or $x + 2 = 0$
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Problem 2: $x(2x + 37) = -80$

Step1: Expand left-hand side

$2x^2 + 37x = -80$

Step2: Rearrange to standard quadratic form

$2x^2 + 37x + 80 = 0$

Step3: Factor the quadratic

$(2x + 5)(x + 16) = 0$

Step4: Solve for $x$

Set each factor to 0: $2x + 5 = 0$ or $x + 16 = 0$
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Problem 3: $3x^5 = 20x^4 + 32x^3$

Step1: Rearrange all terms to left-hand side

$3x^5 - 20x^4 - 32x^3 = 0$

Step2: Factor out common term $x^3$

$x^3(3x^2 - 20x - 32) = 0$

Step3: Factor the quadratic

$x^3(3x + 4)(x - 8) = 0$

Step4: Solve for $x$

Set each factor to 0: $x^3 = 0$, $3x + 4 = 0$, or $x - 8 = 0$
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Problem 4: $6x^3 + 3x^2 - 30x = 0$

Step1: Factor out common term $3x$

$3x(2x^2 + x - 10) = 0$

Step2: Factor the quadratic

$3x(2x + 5)(x - 2) = 0$

Step3: Solve for $x$

Set each factor to 0: $3x = 0$, $2x + 5 = 0$, or $x - 2 = 0$

Answer:

1. $x = 9$ or $x = -2$
2. $x = -\frac{5}{2}$ or $x = -16$
3. $x = 0$, $x = -\frac{4}{3}$, or $x = 8$
4. $x = 0$, $x = -\frac{5}{2}$, or $x = 2$