X2 - X - 2(X + 1)(X - 2)===============(X + 1) is a factor of the above polynomial.
(x + 1) and (x + 2) are monomial factors of the polynomial x2 + 3x + 2 (x + 1) and (x + 3) are monomial factors of the polynomial x2 + 4x + 3 (x + 1) is a common monomial factor of the polynomials x2 + 3x + 2 and x2 + 4x + 3
x2+10x-24 = (x-2)(x+12)
Since the polynomial has three zeroes it has a degree of 3. So we have:(x - -5)[x - (2 + i)][x - (2 - i)]= (x + 5)[(x - 2) - i][(x - 2) + i]= (x + 5)[(x - 2)2 - i2]= (x + 5)[x2 - 2(x)(2) + 22 - (-1)]= (x + 5)(x2 - 4x + 4 + 1)= (x + 5)(x2 - 4x + 5)= x(x2 - 4x + 5) + 5(x2 - 4x + 5)= x3 - 4x2 + 5x + 5x2 - 20x + 25= x3 + x2 - 15x + 25Thus,P(x) = x3 + x2 - 15x + 25
(x + 5)(x + 2)
X2 - X - 2(X + 1)(X - 2)===============(X + 1) is a factor of the above polynomial.
(x-2)(x+2)
If you mean: x2+3x+2 then it is (x+1)(x+2) when factored
(x + 1) and (x + 2) are monomial factors of the polynomial x2 + 3x + 2 (x + 1) and (x + 3) are monomial factors of the polynomial x2 + 4x + 3 (x + 1) is a common monomial factor of the polynomials x2 + 3x + 2 and x2 + 4x + 3
(x + 9)(x + 2)
x2+5x-14 = (x+7)(x-2) when factored
x2+10x-24 = (x-2)(x+12)
(x+4)(x-2)
(x-2)(x-3)
Since the polynomial has three zeroes it has a degree of 3. So we have:(x - -5)[x - (2 + i)][x - (2 - i)]= (x + 5)[(x - 2) - i][(x - 2) + i]= (x + 5)[(x - 2)2 - i2]= (x + 5)[x2 - 2(x)(2) + 22 - (-1)]= (x + 5)(x2 - 4x + 4 + 1)= (x + 5)(x2 - 4x + 5)= x(x2 - 4x + 5) + 5(x2 - 4x + 5)= x3 - 4x2 + 5x + 5x2 - 20x + 25= x3 + x2 - 15x + 25Thus,P(x) = x3 + x2 - 15x + 25
(x + 5)(x + 2)
x^2 + 7x -18 = (x - 2)(x + 9)