Show that the given values of c are zeros of p(x) and find all other zeros of p(x). p(x)=3x^4-8x^3-14x^2+31x+6. C=-2,3
To show that c = -2 is a zero of p(x), we substitute x = -2 into p(x) and see if the result is zero:
p(-2) = 3(-2)^4 - 8(-2)^3 - 14(-2)^2 + 31(-2) + 6
= 3(16) + 8(8) - 14(4) - 62 + 6
= 48 + 64 - 56 - 62 + 6
= 0
Therefore, c = -2 is a zero of p(x).
Similarly, to show that c = 3 is a zero of p(x), we substitute x = 3 into p(x) and see if the result is zero:
p(3) = 3(3)^4 - 8(3)^3 - 14(3)^2 + 31(3) + 6
= 3(81) - 8(27) - 14(9) + 93 + 6
= 243 - 216 - 126 + 93 + 6
= 0
Therefore, c = 3 is a zero of p(x).
To find all other zeros of p(x), we can use polynomial long division or synthetic division to factor out the zeros we have already found. Let's use synthetic division:
-2 │ 3 -8 -14 31 6
-6 28 -28 -6
-----------------------
3 -14 14 3 0
Now, we have a new polynomial: 3x^3 - 14x^2 + 14x + 3. Let's continue dividing:
3 │ 3 -14 14 3
9 -15 -3
-----------------------
3 -5 -1 0
Now, we have another new polynomial: 3x^2 - 5x - 1. Let's divide again:
3 │ 3 -5 -1
9 12
-----------------
3 4 11
The remainder is 11, which means the polynomial 3x^2 - 5x - 1 cannot be further factored with integer coefficients. Therefore, the other zeros of p(x) are approximate values.
From the synthetic division, we have the factored form: (x + 2)(x - 3)(3x^2 - 5x - 1) = 0
To find the other zeros, we need to solve the quadratic equation 3x^2 - 5x - 1 = 0. We can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
For the quadratic equation 3x^2 - 5x - 1 = 0, a = 3, b = -5, and c = -1.
x = (-(-5) ± √((-5)^2 - 4(3)(-1)))/(2(3))
x = (5 ± √(25 + 12))/6
x = (5 ± √37)/6
Therefore, the other zeros of p(x) are approximately:
x ≈ (5 + √37)/6
x ≈ (5 - √37)/6