To say the degrees of the following polynomials 3x5 2x2 x

To say the degrees of the following polynomials; 3x^5 - 2x^2 + x -1; x^3 - 2x^2 + x - 1; 5xyz - x^2 + y - yz + 7. Let p(x) = 2x^2 + x + 1, l_1(x) = 1/2x(x - 1), l_2(x) = - (x - 1)(x + 1), l_3 = 1/2x(x + 1). Evaluate the values: p(-1), p(0) and p(1); Evaluate the values of l_1(x), l_2(x) and l_3(x) at the points: x = - 1, 0, 1; To prove p(x) = p(-1)l_1(x) + p(0)l_2(x) + p(1)l_3(x) Let f(x) = a_2x^2 + a_1x + a_0, g(x) = b_3x^3 + b_2x^2 + b_1x + b_0. Compute the product f(x)g(x). To prove that the set P_3[x] is closed under the polynomial addition and scalar multiplication. Write out the coordinate vectors of the following polynomials in terms of standard bases in decreasing order. 2x^2 + x - 1; (x - 1)(x + 1); x^3 - 2x^2 + x - 1, Given (3, 1, 2, 0) and (2, 1, -1, 3) to be coordinate vectors of two polynomials with respect to the standard bases in decreasing order, write out the general forms of the polynomials. Let p(x) = 2x^2 + x - 1 and q(x) = x^3 - 2x^2 + x - 1 be two polynomial in p_4[x]. To give the coordinate vectors of p(x) + q(x) and 3p(x); Check the results by their coordinate vector operation (the basis is in decreasing order). x^2 - 3x + 2; 2x^2 - 6x + 4; x^2 - 2x - 1 x^3 - 6x^2 + 11x - 6. Let L_1(x) = 1/2x(x - 1), L_2(x) = - (x + 1)(x - 1), L_3(x) = 1/2x(x + 1). Evaluate the values of the three polynomials; L_1(x), L_2(x), L_3(x) at the point: x = -1, 0, 1; Let c_1L_1(x) + c_2L_2(x) + c_3L_3(x) = 0 for all x in R. Then to prove that c_1 = c_2 = c_3 = 0; L_1(x), L_2(x), L_3(x), x^2 - x + 1; 2x^2 - 1; x^2 - 1.

Solution

1) f(x) = 3x^5 -2x^2 +x -1 ; x^3 -2x^2+x -1 ; 5xyz -x^2 +y -yz +7

Degree of  3x^5 -2x^2 +x -1 = highest degree = 5

Degree of  x^3 -2x^2+x -1 = highest degree = 3

Degree of   5xyz -x^2 +y -yz +7 = 3

2) p(x) = 2x^2 + x +1 ; l1(x) = (1/2)x(x-1) ; l2(x) = -(x-1)(x+1) , l3(x) = (1/2)x(x+1)

p(-1) = 2(-1)^2 + (-1) +1 = 2-1 +1 = 2

p(0) = 2(0) -(0) +1 = 1

p(1) = 2(1)^2 +1 +1 = 4