Expand (x+y)4 and (s+y)3. Solve the following quadratic equations using factoring: 1) 3x2 - 27 = 0, 2) 2x2…

Question

Expand (x+y)4 and (s+y)3. Solve the following quadratic equations using factoring: 1) 3x2 - 27 = 0, 2) 2x2 + x − 300 = 0

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to the quadratic equations:

1) Solve $3x^2 - 27 = 0$

Factor out the common term, which is 3.
$3(x^2 - 9) = 0$
Recognize the difference of squares, $x^2 - 9 = x^2 - 3^2$ .
$3(x - 3)(x + 3) = 0$
Set each factor containing $x$ to zero and solve for $x$ .
$x - 3 = 0 \implies x = 3$ $x + 3 = 0 \implies x = -3$
The solutions are $\boxed{x = 3, -3}$ .

2) Solve $2x^2 + x - 300 = 0$
Find two numbers that multiply to $2 \times (-300) = -600$ and add up to the middle coefficient, which is 1.
The numbers are 25 and -24.
$25 \times (-24) = -600$ $25 + (-24) = 1$
Rewrite the middle term ( $x$ ) using these two numbers.
$2x^2 + 25x - 24x - 300 = 0$
Factor by grouping the terms.
$x(2x + 25) - 12(2x + 25) = 0$
Factor out the common binomial term $(2x + 25)$ .
$(2x + 25)(x - 12) = 0$
Set each factor to zero and solve for $x$ .
$2x + 25 = 0 \implies 2x = -25 \implies x = -\frac{25}{2}$ $x - 12 = 0 \implies x = 12$
The solutions are $\boxed{x = -\frac{25}{2}, 12}$ .

Expand (x+y)4 and (s+y)3. Solve the following quadratic equations using factoring: 1) 3x2 − 27 = 0, 2) 2x2 + x − 300 = 0