Math help!?

Question

I need help. The problem says:

Solve without graphing, show all work.

1. x^3+7x^2-8x = 0

2. 2x^3+9x^2= 18x

Can someone please help? Thank you.

Answer 1

Common factor first

x³ + 7x² - 8x = 0

x(x² + 7x - 8) = 0

the first term x² = 1

The middle term inside the parenthesis is + 7x

Find the sum of the middle term

multiply the first term 1 times the last term 8 equals 8 and factor

factors of 8

1 x 8. . .<=. .use these factors
2 x 4

+ 8 and - 1 satisfy the sum of the middle term

- 1 = - x

insert + 8x and - x into the equation

x(x² + 7x - 8) = 0

Group factor

x(x² + 8x - x - 8) = 0

x[x(x + 8) - 1(x + 8) = 0

x(x - 1)(x + 8) = 0

- - - - - -

Roots

x = 0

- - - - - -

Roots

x - 1 = 0

x - `1 + 1 = 0 + 1

x = 1

- - - - - - -

Roots

x + 8 = 0

x + 8 - 8 = 0 - 8

x = - 8

- - - - - - - s-

Answer 2

For both problems, just divide through by x first. When we do that, for problem #1 we get this:

(x³ + 7 x² - 8 x) / x = x² + 7 x - 8.

x² + 7 x - 8 is factorable into (x - 1)(x + 8).

To find possible zeros, set all the factors equal to zero and then solve for x:

x = 0, since x is a factor, because we divided through by x first.
x - 1 = 0 ----> x = 1
x + 8 = 0 ----> x = -8.

So, x = 0, x = 1, or x = -8 are the answers for this problem.

For problem #2, first move 18x to the left side of the equation. When we do that, it becomes -18x, resulting in this final form:

2 x³ + 9 x² - 18 x = 0.

Then, when we divide through by x, we get this quotient:

2 x² + 9 x - 18, which is factorable into (2 x - 3)(x + 6). Now, we set all these factors equal to 0 and solve for x:

x = 0 for the same reason as problem #1
2 x - 3 = 0 ----> 2 x = 3 ----> x = 3/2
x + 6 = 0 ----> x = -6.

So, x = 0, x = 3/2, or x = -6 are the answers for problem #2.

Answer 3

1. Factor out an x: x(x^2+7x-8)=0
then factor the parenthases: x(x+8)(x-1)=0
then set each factor equal to 0
x=0 x+8=0 x-1=0
then solve each equation for x
x=0 x=-8 x=1

2. Set equation equal to 0: 2x^3+9x^2-18x
then follow the same steps as number one:
x(2x^2+9x-18)=0
x(2x-3)(x+6)=0
x=0 2x-3=0 x+6=0
x=0 x=3/2 or 1.5 x=-6

Answer 4

x^3+7x^2-8x = 0
x(x^2+7x-8)=0 (take out x which is common)
x(x^2-x+8x-8) = 0 (7x can be rewritten as -x+8x)
x(x(x-1)+8(x-1))=0 (x is common from first 2 terms and 8 is common from last 2 terms)
x(x-1)(x+8)=0 (x-1) is common which leaves behind (x+8)
so answers are
x = 0, x = 1, x = -8
-----------------------------------------------------------------------------------
2x^3+9x^2= 18x
x(2x^2+9x-18) = 0 (bring 18x on LHS and so the sign changes to -ve and take out x which is common)
x(2x^2+12x-3x-18)=0 (9x can be rewritten as 12x-3x)
x(2x(x+6)-3(x+6))=0 (2x is common from first 2 terms and -3 is common from last 2 terms)
x (x+6)(2x-3) = 0 (x+6 is common which leaves (2x-3))
x = 0, x = -6, x = 3/2

Answer 5

x^3+7x^2-8x=0
=>x(x^2+7x-8)=0
=>x{x^2-x+8x-8)=0
=>x{x(x-1)+8(x-1)}=0
=>x(x-1)(x+8)=0
=>x=0,1 or -8
2.2x^3+9x^2=18x
=>2x^3+9x^2-18x=0
=>x(2x^2+9x-18)=0
=>x(2x^2+12x-3x-18)=0
=> x{2x(x+6)-3(x+6)}=0
=> x(x+6)(2x-3)
Therefore either x=0 or -6 or 3/2

Answer 6

1. Answer: 1, -8, 0
2. Answer: -6, 3/2, 0

Just trust me on this one.