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You can use the BLANK of a polynomial to help you find its factors or roots?
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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
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.
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
How can you find out how many solutions an equation has?
By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.
In a cubic polynomial 2xxx-5xx-14x 8 find the sum of the zeros?
2x^3 - 5x^2 - 14x + 8 Let P(x) represents the cubic polynomial. We can find the sum of x-values which make P(x) = 0, (the sum of the roots of the equation) P(x) = 2x^3 - 5x^2 - 14x + 8 P(x) = 0 2x^3 - 5x^2 - 14x + 8 = 0 Since the degree of this polynomial is odd, then the sum of the roots is -[a(n - 1)/an], where a(n-1) is -5 and an is 2. So we have, -[a(n - 1)/an] = -(-5/2) = 5/2 Thus the sum of the roots is 5/2.
You can also use of a polynomial to help you find its factors and roots?
graph!
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 - 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 - 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 - c is a factor?
a
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.
What is the rule for this table 52 44 47 39 40 blank 36 28?
Given any number for the blank it is easy to find a rule based on a polynomial of order 7 such that the seven numbers are as listed in the question and the blank are, in order, the numbers generated by the polynomial. There are also non-polynomial solutions.The solution, based on a polynomial of order 6 is:Un = 0.09187n6 - 2.46964n5 + 25.93353n4 - 134.62797n3 + 359.54602n2 - 462.47380n + 266 for n = 1, 2, 3, ...Accordingly, the blank is 47.
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).
You can also use the of a polynomial to help you find its factors?
graph apex xD
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
