Question

PLEASE HELP ASAP!!!

Answer 1

Part 1:

$(x+1)^2-3(x+2) \\ \\ =x^2+2x+1-3x-6 \\ \\ =x^2-x-5$

option b.


Part 2:
$(5x^3y^{-3})(-6x^4y)\\ \\=5(-6)x^{3+4}y^{-3+1}\\ \\=-30x^7y^{-2}\\ \\=\frac{-30x^7}{y^2}$

option d


Part 3:
$\frac{8+x}{2} - \frac{x+3}{3} =4$

Multiply through by the LCM of 2 and 3 (i.e. 6)
$6\left(\frac{8+x}{2}\right) - 6\left(\frac{x+3}{3}\right) =6(4) \\ \\ 3(8+x)-2(x+3)=24 \\ \\ 24+3x-2x-6=24 \\ \\ x+18=24 \\ \\ x=24-18=6$

option b


Part 4
x^2-3x-8=x+4

$x^2-3x-8=x+4 \\ \\ x^2-3x-x-8-4=0 \\ \\ x^2-4x-12=0 \\ \\ (x-6)(x+2)=0 \\ \\ x=6 \ or \ -2$

option d


Part 5
Joseph is solving the equation x^2+8x-4=0 using the technique completing the square.
x^2+8x-4=0
x^2+8x+?=4+?

In completeing the square method, you divide the coeficient of x by 2, square the result and add to both sides of the equation.

Therefore, he should add 16 to both sides because (8/2)^2=16.

option d.