Polynomials/linear factors hw help?
The problem states: Use the Table feature of your graphing calculator to complete the table of values for p(x) = x^4 - 2x^3 - 9x^2 + 2x + 8. Use your results to factor p(x) completely. Justify your answer. So this is the info I have to use from the table: x | -4 , -3 , -2 , -1 , 0 , 1 , 2 , 3 , 4 , 5 p(x) | 240 , 56 , 0 , 0 , 8 , 0 , -24 , -40 , 0 , 168 Can someone please help? I'm not sure how to go about solving this. I would assume that if I factor completely and the leading term is x^4 then four linear factors would equal to the polynomial. But how do I find these four linear factors with the information in the table?
What is the linear factor of 2x² -x -6?
2x^2 - x - 6 Find two factors of -12, when added or subtracted, give -1 as an answer. In this case, -4 + 3 = -1 and -4 * 3 = -12 Rewrite -x as -4x + 3x: 2x^2 - 4x + 3x - 6 (2x^2 - 4x) + (3x-6) Factor out a GCF of 2x and 3, respectively out of each group: 2x(x-2) + 3(x-2) Factor by grouping --- a(m+n) + b(m+n) = (a+b)(m+n). Applying this: (2x+3)(x-2) Therefore, the answer is C.
How do you factor 12x^2 - 4 into linear factors? The term linear factors is important.
I would do it this way[math]12x^2–4[/math][math]=4(3x^2–1)[/math]Now the part inside the parenthesis is not exactly a perfect square but I can force it[math]=4(\sqrt{3}x+1)(\sqrt{3}x-1)[/math]There we go, linear factors since the expressions have [math]1[/math] as the power of [math]x[/math]. That’s the definition of a factor to be linear.
Which of the following is a linear factor of 2x2 − x − 6?
2x² - x - 6 = 0 2x² + 3x - 4x - 6 = 0 x(2x + 3) - 2(2x + 3) = 0 (x - 2)(2x + 3) = 0 C. 2x + 3
2 questions about factoring, please help?
Write a linear factorization of the function. f(x) = x4 + 64x2 f(x) = x2(8x + i)(8x - i) f(x) = x2(x + 8i)2 f(x) = x2(x + 8i)(x - 8i) f(x) = x2(8x + i)2 Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 8, -14, and 3 + 9i f(x) = x4 - 11x3 + 72x2 - 606x + 10,080 f(x) = x4 - 303x2 + 1212x - 10,080 f(x) = x4 - 11x3 - 72x2 + 606x - 10,080 f(x) = x4 - 58x2 + 1212x - 10,080
Show that x^3 - 8 has only one linear factor over the real number field?
It is (x-2)(x^2+2x+4) and second factor can not be factorised. Its discriminant being -ve.
Does there exist a linear-time factoring algorithm?
Not even close. There are simple algorithms that are O( sqrt(N) ). And there are faster factoring algorithms. Interestingly, we do not know yet what the fastest factoring algorithm is. And this is an important part of number theory, because if we can factor large numbers fast, some of our encryption techniques, such as the RSA Algorithm, will break.For example, suppose we generate two 1000 digit primes, P1 and P2. You'd need to know some Discrete Math to be able to find such primes, but it can be done quickly.Now compute C = P1 * P2 and give C to any organization in the world.C will have about 2000 digits, which is well over 6000 bits.With out current knowledge of factoring, the Sun would burn out before someone could factor C.If you figure out a way to factor C quickly, let me know. We'll make a ton of money selling the algorithm to the National Security Agency. But don't tell anyone else, otherwise 1) No Money, and 2) You'd be compromising the security of the United States.
What are the linear factors of a polynomial?
A polynomial is a function of a variable which consists of a combination of terms with different powers of the variable (e.g. [math]x^3 + 3.x^2 - 4.x - 12[/math]). One way of finding roots for the variable is factoring the expression. Thus, for the example in consideration,[math]f(x) = x^3 + 3.x^2 - 4.x - 12[/math][math]f(x) = (x^2-4)(x+3)[/math]Roots are found typically, when the expression is equal to zero. With this factorization, the variable can be solved.In the example above, the expression [math]f(x)[/math] is solved for a multiplication of 2 factors. One is a linear factor [math](x+3)[/math], which leads to a single constant solution (-3) and the other is a quadratic factor [math](x^2-4)[/math], which leads to 2 solutions (+2 or -2).Simply put, a linear factor of a polynomial is a factor of which the power of the variable is 1 (e.g. [math](x+3)[/math] ). Such a factor will give a constant as the solution for the variable.
Factoring into linear or irreducible quadratic factors...?
Special Products of factorizations • (x + y)2 = x2 + 2xy + y2 • (x − y)2 = x2 − 2xy + y2 • (x + y)3 = x3 + 3x2y + 3xy2 + y3 • (x − y)3 = x3 − 3x2y + 3xy2 − y3 • (x + y)(x − y) = x2 − y2 • (x + y)(x2 − xy + y2) = x3 + y3 • (x − y)(x2 + xy + y2) = x3 − y3 1.x^3 – 27 (x-3)(x^2+3x+9)
Factoring polynomials into linear factors of (poly mod n)?
Hello could someone please help me do these 3 equations by factoring these polynomials into linear factors of (poly mod n)? 1) 2x^2 - 3x - 2 into linear factors (poly mod 7) 2) x^2 + 1 into linear factors (poly mod 17) 3) x^3 + 4x^2 + 3x + 6 into linear factors (poly mod 7) thank you for the help in advance !