(x2 + 1)(x - 3)
x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)
x3 - 3x2 + x - 3 = (x - 3)(x2 + 1)
3x2 - 2x + 3
It is a polynomial if the square root is in a coefficient but not if it is applied to the variable. A polynomial can have only integer powers of the variable. Thus: sqrt(2)*x3 + 4*x + 3 is a polynomial expression but 2*x3 + 4*sqrt(x) + 3 is not.
f(x)=x3-3x2-5x+39=(x+3)(x2-6x+13) It has three roots. One of which is x=-3. Using the quadratic equation: x = (6 +/- √(-16))/2 x = (6 +/- 4i)/2 = (3 +/- 2i) so, x=-3, x=3+2i, or x=3-2i
x3 - 3x2 + x - 3 = (x2 +1)( x - 3)
x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)
Answer this ques Which polynomial represents the sum below?(-x3 + 3x2 + 3) + (3x2 + x + 4)tion…
x3 - 3x2 + x - 3 = (x - 3)(x2 + 1)
3x3 -3x2 - 48x - 60 = 0 simply factor out a 3... 3(x3 - x2 - 16x - 20) = 0
x3 + 3x2 - 4x -12 x2(x + 3) - 4(x + 3) (x2 - 4)(x+3)
(x3)'=3x2(x3)''=(3x2)'=6x
3x2 - 2x + 3
To factor the polynomial ( x^3 - 3x^2 + x - 3 ), we first apply the Rational Root Theorem and test possible rational roots. After testing, we find that ( x = 3 ) is a root. Using synthetic division, we divide the polynomial by ( (x - 3) ) to obtain ( x^2 + 1 ). Thus, the complete factorization is ( (x - 3)(x^2 + 1) ).
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.
3x2-9x-30 =3x2+6x-15x-30 =3x(x+2)-15(x+2) =(3x-15)(x+2) OR 3(x - 5)(x+2)
It is -2*(3x2 - 7)*(x3 - 7x)-3