In Exercises 12–13, subtract the polynomials. 12. (7x – 6x2 + 5x – 11) – (-8x³ + 4x2 – 6x – 3) 13. (4x*y² – 7x°y – 4) – (6x³y? - 3x'y + 4) 14. Subtract -2x³ – x²y + xy? + 7y3 from x + 4x?y – y.
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- The results obtained as a result of velocity measurements made from the moment a rocket was launched are shown in the table below.is seen. Equation connecting rocket velocity to time V(t) = a1t2 + a2t + a3 5 ≤ t ≤ 12 Since it is known that there is a polynomial in the form, calculate the value of the coefficients a1, a2, a3 in the equation by Gauss elimination method.I don't understand how g(x)=(x-2)^4 is a product of irreducible polynomials over Z3. Please explain. Thank you.Show that the cubic polynomial p(x) = ax3 + bx2 + cx + d has exactly one point of inflection (x0, y0), where value of x0 and y0 is attached. Use these formulas to find the point of inflection of p(x) = x3 − 3x2 + 2.
- Given two polynomials in from of x^3+x^2+x+c =0 x^3+x^2+x+c =0 how do you find their intersection using matricies?Find the fifth Taylor polynomial of x3 - 7x2 + 8 at x = 0.Find a third-degree polynomial p(x) that is tangent to the line y = 14x − 13 at the point (1, 1), and tangent to the line y = −2x − 5 at the point (−1, −3).
- what is a polynomial of the degree 4 with the interger coefficients with zeros of 1+3i and 2 (with 2 as a zero of miltiplicity two)? Completely expand (multiply out) the answer.Find the Taylor polynomial centered at C=4 f(x)=x1/8, n=3Find a fourth degree polynomial with real coefficients that has zeros of -3, 2, i that passes through (-2, 100).
- Does there exist a polynomial of degree 4 with real coefficients that has zeros i, 2i, 3i and 4i? Why or why not?I am trying to find out if the polynomial 12x^4+8x^3+9x^2+4x+4 is reducible or irreducible over Q. We can't apply Eisenstein's criterion here. But we can try this:12x^4+8x^3+9x^2+4x+4 = x^4+(2/3)x^3+(3/4)x^2+(1/3)x+(1/3) x^4+(2/3)x^3+(3/4)x^2+(1/3)x+(1/3)=(x^2+ax+b)(x^2+cx+d) = x^4+(a+c)x^3+(b+d+ac)x^2+(ad+dc)x+bd We get the equation system: a+c=(2/3)b+d+ac=(3/4)ab+bc=(1/3)bd=(1/3) I get the following result:b=(1/2)d=(2/3)c=(2+sqrt(19))/6 and (2-sqrt(19))/6a=(2/3)-((2+sqrt(19))/6) and (2/3)-((2-sqrt(19))/6) The expression in a contains an irrational number, sqrt(19), this means that the polynomial is irreversible over Q. Is this correct?Express p (x) = x3 as a Taylor polynomial about a = 1/2