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What else can I help you with?
You can use the BLANK of a polynomial to help you find its factors or roots?
Graph factor
You can also use the of a polynomial to help you find its factors?
graph apex xD
The polynomial given roots?
Do mean find the polynomial given its roots ? If so the answer is (x -r1)(x-r2)...(x-rn) where r1,r2,.. rn is the given list roots.
How do you find a degree of a graphed polynomial?
Factors
To find the factors of a polynomial from its graph follow this rule If the number is a root of a polynomial then x - b is a factor?
a
To find the factors of a polynomial from its graph follow this rule If the number is a root of a polynomial then x - a is a factor?
B
To find the factors of a polynomial from its graph follow this rule If the number is a root of a polynomial then x - c is a factor?
a
How can one find all rational roots of a polynomial equation?
To find all rational roots of a polynomial equation, you can use the Rational Root Theorem. This theorem states that any rational root of a polynomial equation in the form of (anxn an-1xn-1 ... a1x a0 0) must be a factor of the constant term (a0) divided by a factor of the leading coefficient (an). By testing these possible rational roots using synthetic division or polynomial long division, you can determine which ones are actual roots of the equation.
Find the roots of the polynomial given 4x2 - 3x - 5?
No integer roots. Quadratic formula gives 1.55 and -0.81 to the nearest hundredth.
What are the roots of the polynomial x2 plus 3x plus 5?
You can find the roots with the quadratic equation (a = 1, b = 3, c = -5).
Can all quadratic equations be solved?
Well, that depends on what you mean "solve by factoring." For any quadratic equation, it is possible to factor the quadratic, and then the roots can be recovered from the factors. So in the very weak sense that every quadratic can be solved by a method that involves getting the factors and recovering the roots from them, all quadratic equations can be solved by factoring. However, in most cases, the only way of factoring the quadratic in the first place is to first find out what its roots are, and then use the roots to factor the quadratic (any quadratic polynomial can be factored as k(x - r)(x - s), where k is the leading coefficient of the polynomial and r and s are its two roots), in which case trying to recover the roots from the factors is redundant (since you had to know what the roots were to get the factors in the first place). So to really count as solving by factoring, it makes sense to require that the solution method obtains the factors by means that _don't_ require already knowing the roots of the polynomial. And in this sense, most quadratic equations are not solvable through factoring.
Can you find a third degree polynomial equation with rational coefficients that has the given numbers as roots 3i and 7?
Yes, easily. Even though the question did not ask what the polynomial was, only if I could find it, here is how you would find the polynomial: Since the coefficients are rational, the complex (or imaginary) roots must form a conjugate pair. That is to say, the two complex roots are + 3i and -3i. The third root is 7. So the polynomial, in factorised form, is (x - 3i)(x + 3i)(x - 7) = (x2 + 9)(x - 7) = x3 - 7x2 + 9x - 63
