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3x2 - 2 is a polynomial of order 2. Therefore, dividing it by (x + 1) will result in a polynomial of order 1. Suppose the quotient is ax + b (where a is non-zero), and with the remainder c. Thus 3x2 - 2 = (x + 1)*(ax + b) + c = ax2 + ax + bx + b + c = ax2 + (a + b)x + (b + c) Comparing coefficients: 3 = a 0 = a + b => 0 = 3 + b => b = -3 -2 = b + c => -2 = -3 + c => c = 1 Therefore, (3x2 - 2)/(x + 1) = 3x - 3 = 3*(x - 1) and a remainder of 1.
2x4 - 9x3 + 13x2 - 15x + 9 = 2x4 - 6x3 - 3x3 + 9x2 + 4x2 - 12x - 3x + 9 = 2x3(x - 3) - 3x2(x - 3) + 4x(x - 3) - 3(x - 3) = (x - 3)*(2x3 - 3x2 + 4x - 3) So the quotient is (2x3 - 3x2 + 4x - 3) and the remainder is 0.
It is of the 2nd degree.
No, it is not. f(x) = 2x + 3 and g(x) = 3x2 are polynomials but f(x)/g(x) is not a polynomial.
the solutions to this equation are -1,+1 and -3. you can solve this equation by using the polynomial long division method. we basically want to factorize this and polynomial and equate its factors to zero and obtain the roots of the equation. By hit and trial , it clear that x=1 i.e is a root of this equation. So (x-1) should be a factor of the given polynomial (LHS). Divide the polynomial by x-1 using long division method and you will get the quotient as x2+4x+3 and remainder would be 0 ( it should be 0 as we are dividing the polynomial with its factor. Eg when 8 is divided by any of its factor like 4,2 .. remainder is always zero ) Now, we can write the given polynomial as product of its factors as x3+3x2-x-3 = (x-1)(x2+4x+3) =(x-1)(x+1)(x+3) [by splitting middle term method] so the solutions for the given polynomial are obtained when RHS = 0, Hence x=-1 , X = +1, x=-3 are the solutions for this equation.
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Answer this ques Which polynomial represents the sum below?(-x3 + 3x2 + 3) + (3x2 + x + 4)tion…
The degree of this polynomial is 2.
3x2 - 2 is a polynomial of order 2. Therefore, dividing it by (x + 1) will result in a polynomial of order 1. Suppose the quotient is ax + b (where a is non-zero), and with the remainder c. Thus 3x2 - 2 = (x + 1)*(ax + b) + c = ax2 + ax + bx + b + c = ax2 + (a + b)x + (b + c) Comparing coefficients: 3 = a 0 = a + b => 0 = 3 + b => b = -3 -2 = b + c => -2 = -3 + c => c = 1 Therefore, (3x2 - 2)/(x + 1) = 3x - 3 = 3*(x - 1) and a remainder of 1.
3x2 - 2x + 3
2x4 - 9x3 + 13x2 - 15x + 9 = 2x4 - 6x3 - 3x3 + 9x2 + 4x2 - 12x - 3x + 9 = 2x3(x - 3) - 3x2(x - 3) + 4x(x - 3) - 3(x - 3) = (x - 3)*(2x3 - 3x2 + 4x - 3) So the quotient is (2x3 - 3x2 + 4x - 3) and the remainder is 0.
It is of the 2nd degree.
2x^3 - 3x^2 + 4x - 3
That depends on whether or not 2x is a plus or a minus
3x2 - 5x - 2 can be factored into (3x + 1) (x - 2)
x3 - 3x2 + x - 3 = (x2 +1)( x - 3)