(x^2-2x-15)(x-4)=0 find the roots to solve?

Answer 1

the given equation is

(x^2 - 2x - 15)(x - 4) = 0
or, (x^2 - 5x + 3x - 15)(x - 4) = 0
or, (x(x-5) + 3(x-5))(x-4) = 0
or, (x-5)(x+3)(x-4) = 0

so the solution is x = 5,-3, 4

Hope this was helpful.

Answer 2

(x^2 - 2x - 15)(x - 4) = 0
(x - 5)(x + 3)(x - 4) = 0

x = 5, -3, 4

Those are the roots

Answer 3

Solve the x-variables: (x^2-2x-15)(x-4) = 0

First: factor the 1st set of parenthesis. Multiply the 1st & 3rd coefficient to get "-15." Find two numbers that give you "-15" when multiplied & "- 2" (2nd/middle coefficient) when added/subtracted. The numbers are (- 5 & 3).

Sec: rewrite the expression with the new middle coefficients.

x^2 - 5x + 3x - 15

*With 4 terms - group "like" terms & factor both sets of parenthesis.

(x^2 - 5x) + (3x - 15)
x(x - 5) + 3(x - 5)
(x-5)(x+3)

Now,you have....

(x-5)(x+3)(x-4) = 0

Third: solve the x-variables > set all parenthesis to 0.

a. x - 5 = 0

*Isolate "x" on one side - add 5 with both sides.

x - 5 + 5 = 0 + 5
x = 5

b. x + 3 = 0

*Isolate "x" on one side - subtract 3 from both sides.

x + 3 - 3 = 0 - 3
x = - 3

c. x - 4 = 0

*Isolate "x" on one side - add 4 with both sides.

c. x - 4 = 0

*Isolate "x" on one side - add 4 with both sides.

x - 4 + 4 = 0 + 4
x = 4

Solutions: - 3, 4, 5

Answer 4

(x² - 2x - 15)(x - 4)

(x - 5)(x + 3)(x - 4)

- - - - - -

Roots
x - 5 = 0

x - 5 + 5 = 0 + 5

x = 5

- - - - - - -

Roots

x + 3 = 0

x + 3 - 3 + 0 - 3

x = - 3

- - - - - -- -

Roots

x - 4 = 0

x - 4 + 4 = 0 + 4

x = 4

- - - - - - - - - -s-

Answer 5

(x² - 2x - 15) . (x - 4)= 0

(x - 5) . (x + 3) . (x - 4) = 0

x = - 3, 4 or 5

Answer 6

= (x-5)(x+3)(x-4)=0
x=5
x=-3
x=4

Answer 7

(x+3)(x-5)(x-4)=0

x1=-3, x2=5, x3=4