Solve the equation 6x^3+5x^2–16x–15=0 given that one of the roots is an integer.?

f(x) = 6x^3+5x^2-16x -15...(1). One of the zeros of f(x) is an integer. Try x = -1.

f(-1) = -6 +5+16-15 = 0. Then (x+1) is a factor of f(x). Put g(x) = 6x^2+ax-15. Then

(x+1)g(x) = f(x), ie., 6x^3+ax^2-15x+6x^2+ax-15 = 6x^3+(a+6)x^2+(a-15)x-15...(2).

Equating coefficients of like powers of x in (1) and (2) gives a+6=5 and a-15 = -16

which are both satisfied when a = -1. Therefore g(x) = 6a^2-x-15 = (2x+3)(3x-5).

Then zeros of f(x) are -3/2, -1 and 5/3.

6x³ + 5x ²– 16x – 15 = 0

Factoring:

(x + 1) (2x + 3) (3x - 5) = 0

If the product of three terms equals 0, then any one or all terms equals 0.

If x + 1 = 0,

x = - 1

¨¨¨¨¨¨¨¨¨

If 2x + 3 = 0, 

2x = - 3

x = - 3/2

¨¨¨¨¨¨¨¨¨¨¨

If 3x - 5 = 0,

3x = 5

x = 5/3

¨¨¨¨¨¨¨¨¨

Solution Set: {- 1, - 32, 5/3}

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

 6x^3+5x^2–16x–15=0,  6x^3+6x^2-x^2–1x-15x–15=0, 6x^2(x+1)-x(x+1)-15(x+1)=0;  (x+1) (6x^2-x -15)=0

6x^3 + 5x^2 - 16x - 15 = 0

(2x + 3)(x + 1)(3x − 5) = 0

The sum of coeffs of even powers is  E= +5 -15 = - 10

The sum of coeffs of odd powers is O = 6 - 16 = -10

If E+O = 0 the x=1 is a rooot

IF E-O = 0 then x=-1 is a root

In this case x=-1 is a root

After synthetic diviciosn   the poly factorizes into (x+1)(6x^2 - x -15)

The other two roots   are  ( 1 +- sqrt( 1 - 4(6)(-15)) / 12

or  (  1 +-  sqrt(361)) / 12  =   (  1 +-  19)/6 =  -3,  10/3

Possible rational roots:  d(15) / d(6) where d(n) is some divisor of n

15: -15 , -5 , -3 , -1 , 1 , 3 , 5 , 15

6: -6 , -3 , -2 , -1 , 1 , 2 , 3 , 6

+/- 15/6 , +/- 15/3 , +/- 15/2 , +/- 15/1 , +/- 5/6 , +/- 5/3 , +/- 5/2 , +/- 5/1 , +/- 3/6 , +/- 3/3 , +/- 3/2 , +/- 3/1 , +/- 1/6 , +/- 1/3 , +/- 1/2 , +/- 1/1

+/- 2.5 , +/- 5 , +/- 7.5 , +/- 15 , +/- 5/6 , +/- 5/3 , +/- 2.5 +/- 5 , +/- 1/2 , +/- 1 , +/- 1.5 , +/- 3 , +/- 1/6 , +/- 1/3 , +/- 1/2 , +/- 1

We were told that one of our roots was going to be an integer, so let's pick integers, shall we?

-5 , 5 , -15 , 15 , -1 , 1 , -3 , 3

6x^3 + 5x^2 - 16x - 15 = 0

The coefficients are pretty small, so let's pick values that are closer to 0

x = 1 : 6 + 5 - 16 - 15 = 11 - 31 = -20

x = -1 : -6 + 5 + 16 - 15 = 21 - 21 = 0

x = -1 works, which means that x + 1 is a factor

(x + 1) * (ax^2 + bx + c) = 6x^3 + 5x^2 - 16x - 15

ax^3 + bx^2 + cx + ax^2 + bx + c = 6x^3 + 5x^2 - 16x - 15

ax^3 = 6x^3

a = 6

bx^2 + ax^2 = 5x^2

a + b = 5

6 + b = 5

b = -1

cx + bx = -16x

c + b = -16

c - 1 = -16

c = -15

(x + 1) * (6x^2 - x - 15) = 0

6x^2 - x - 15 = 0

x = (1 +/- sqrt(1 + 360)) / 12

x = (1 +/- 19) / 12

x = 20/12 , -18/12

x = 5/3 , -3/2

x = -3/2 , -1 , 5/3

Those are your roots

x = 5/3

3x = 5

3x - 5

x = -3/2

2x = -3

2x + 3 = 0

(x + 1) * (3x - 5) * (2x + 3)

That's the polynomial factored.

Since one root is an integer (thus rational), use the RRT to find this root.

Once found, factor it out, leaving you with a quadratic, which you know how to solve.

Done!

Show your steps here if need be and we can further proceed if required.