Question

Suppose p(x) is a polynomial of smallest possible degree such that: bullet p(x) has rational coefficients bullet p(−3)=p(sqrt7)=p(1−sqrt6)=0 bullet p(−1)=8 determine the value of p(0).

Answer 1

Answer:

p(0) = 28.07

Step-by-step explanation:

Since p(-3)=0, it means x= -3 is a root of the polynomial. And the factor is x-(-3) = x + 3

Also, p(sqrt7)=0, meaning that x= $\sqrt{7}$ is a root of the polynomial. And the factor is x - $\sqrt{7}$

Lastly, p(1-sqrt6)=0, meaning that x= $1-\sqrt{6}$ is a root of the polynomial. And the factor is x - $(1-\sqrt{6})$

The polynomial p(x) = K(x+3)(x-√7)(x-(1-√6))

But p(-1) = 8. That is when x = -1, p(x) = 8. So we use this to find K

$8 = K(-1 + 3)(-1 -\sqrt{7})(-1-(1-\sqrt{6})\\ \\8 = K(2)(-1 -\sqrt{7})(-2 +\sqrt{6}) = K(2)(-3.6457)(0.4495)\\\\8 = -3.2775K\\\\\frac{8}{-3.2775} = K\\\\K = -2.4409$

K ≅ -2.44

$p(x) = -2.44(x + 3)(x -\sqrt{7})(x-(1-\sqrt{6})\\\\p(0) = -2.44(0+3)(0-\sqrt{7})(0-(1-\sqrt{6})\\\\p(0) = -2.44(3)(-2.6457)(1.4495) = 28.0716$

p(0) ≅ 28.07