6) polynomials?

Question

if x - k, where k is integral, is a factor of the polynomial P(x) = 2x^3 - 3x^2 - 12x - 7 find all possible values for k.

Factorise P(x) completely and sketch y= P(x)

Answer 1

First list the possible rational roots for the polynomial

P = -7, Q = 2

All the possible rational roots = P/Q = +/- 7, +/-1, +/- 1/2, +/- 7/2

Then input these x-values into the function to find the zeros (those that output zero)

So the first that output a zero was -1

So divide out (x + 1) which is a factor (either with synthetic division or traditional polynomial division)


(2x^3 - 3x^2 - 12x - 7) / (x + 1) = 2x^2 - 5x - 7

Factor the remaining trinomial normally

P(x) = (x + 1)(2x^2 - 5x - 7)

P(x) = (x + 1)(2x - 7)(x + 1)

The polynomial has

a double root at -1 and another root at 7/2

Plot these two points on the x-axis, then check values to the left of -1, between the roots, and to the right of 7/2 and connect the points with a continous curve.

Answer 2

Trying -1

2*(-1)^3 - 3(-1)^2 - 12*(-1) - 7 = 2*(-1) - 3*1 - 12*(-1) - 7 =
= -2 -3 + 12 - 7 = 0

2x^3 - 3x^2 - 12x - 7 = 2x^3 - 3x^2 - 5x - 7x - 7 =
= 2x^3 - 3x^2 - 5x - 7(x + 1) = 2x^3 + 2x^2 - 5x^2 - 5x - 7(x + 1) =
= 2x^3 + 2x^2 - 5x(x + 1) - 7(x + 1) = 2x^2(x +1) - 5x(x + 1) -
- 7(x + 1) = (2x^2 - 5x - 7)(x + 1)

Let's factorize (2x^2 - 5x - 7)

x1,2 = [5 +- sqrt(25 + 56)]/4 = (5 +- 9)/4

x1 = -4/4 = -1

x2 = 14/4 = 3.5

P(x) = (x + 1)^2 * (x -3.5)

Now, P'(x) = 6x^2 - 6x - 12 = 6(x^2 -x - 2)

Let's find roots:

x1,2 = [1 +- sqrt (1 + 8)]/2 = (1 +- 3)/2

x1 = (1+3) / 2 = 2
x2 = (1 - 3)/2 = -1

Let's determine by P''(x) is the points are local maximum or minimum points

P''(x) = 12x - 6

at x=2 P''(x) = 12*2 - 6 = 24 - 6 = 18 > 0, hence 2 is a local minimum.

at x=-1 P(X) = 12*(-1) - 6 = -12 - 6 = -18 <0, hence -1 is a local maximum.

Answer 3

If P(x) with integral coeffecients has an integral root, it must be a factor of the absolute term.
So the possible integral roots are 1,-1,7,-7
We find that -1 is the only posible value for k for P(-1) = 0
Dividing by x+1 we get
P(x) = (x+1) ( 2 x^2 - 5x -7 ) = (x+1)(2x -7)(x+1)
then sketch it using the intersecting points with x & y axes :
(-1,0) , ( 3.5,0), (0,-7) and some more additional points

Answer 4

you are able to seem on your answer below the type (x^2 + 2px + p^2 + A)*(x^2+ 2x + a million + B)*(x^2 - 2*(p+a million)x + (p+a million)^2 + C) with the situations that p >= a million, A > 0, B > 0 C > 0. figuring out to 0 the coefficients of x^4, x^3, x^2 you get here equations A + B + C = 2*(a million+p+p^2) pA + B - (p+a million)C = -p(p+a million) AB + BC + CA + A(p^2 - 2p - 2) +B*(-2p^2+2p+a million) + C*(p^2+4p+a million) = 0 Edit: I forgot a term +p^2 + (p^2+a million)*(p+a million)^2. subsequently what comes after should be marginally changed.... removing C one is decreased to (2p + a million)*A + (p + 2)*B = (p + a million)*(2p^2 + p + 2) pA^2 + 2(p+a million)AB + B^2 + A*(2p^3+4p^2-2p-2) + B*(-2p^3-2p^2+4p+2) + p*(p+a million)(p^2+4p+a million) = 0. observe that (p+a million)^2-p >0. So we are decreased to understanding whilst a hyperbola and a line, with coefficients looking on p, have rational intersections, certainly one of which being with non unfavorable coordinates. The existence of an intersection element with constructive coordinates seems ok for p sufficiently huge, say p>3. however the rationality area reduces to exhibiting that Q(p) is the sq. of a rational, the place Q is a polynomial of degree 10 with integer coefficients, springing up by using fact the discriminant of a 2nd order equation in A. So probable in straightforward terms Maple or Mathematica human beings have a raffle...

Answer 5

(x+1)(x+1)(2x-7) k=-1; 7/2 isn't an integer but it does satifsy the equation