How to factor x^2+16?
I know that x^2-16 is a perfect square but +16 will not factor. I believe there is a formula similar to the cube root formula but not sure? Maybe I am just over thinking this. Thanks for the help
Polynomial Factoring?????????????????????...
don't try to do it all at once factor the integers, then each varfiable individually 1.GCF of 8x^4, -24x^2 8(x^4, -3x^2) 8x^2(x^2, -3) 2.16p^6q^4, 32p^3q^3, -48pq^2 16(p^6q^4, -2p^3q^3, -3pq^2) 16p(p^5q^4, -2p^2q^3, -3q^2) 16pq^2(p^5q^2, -2p^2q, -3) 3.Factor: 8x^2-4x-20 4(2x^2-x-5) quadratic formula b^2-4ac 1+40 41 +/-sqrt41 rest of q.f. (-b+/-sqrt41)/2 (1+/-sqrt41)/2
What are the prime numbers p such that 16p+1 is a cube of some positive integers?
If [math]p[/math] is a prime such that [math]16p+1[/math] is the cube of some positive integer, there exists a positive integer [math]a[/math] such that 1[math]6p+1=a^3[/math]. Then [math]16p=a^3–1=(a-1)(a^2+a+1)[/math]. Note, however, that [math]a^2+a+1=a(a+1)+1[/math]. Since the product of two any consecutive integers is even, [math]a(a+1)+1[/math] is necessarily odd. Therefore [math](a-1)[/math] is divisible by [math]16[/math], which means that there exists an integer [math]k[/math] such that [math](a-1)=16k[/math].Then [math]16p=16k(a^2+a+1)[/math]. This is equivalent to say that [math]p=k(a^2+a+1)[/math]. Since [math]p[/math] is prime, either [math]k[/math] or [math](a^2+a+1)[/math] is equal to [math]1[/math] - otherwise [math]p[/math] wouldn’t be prime. If [math](a^2+a+1)=1[/math], [math]a=0[/math] or [math]a=-1[/math], but [math]a[/math] is a positive integer. That means [math]k[/math] need to be equal to [math]1[/math]. Hence, [math](a-1)=16[/math] and consequently [math]a=17[/math]. Thus, [math]16p+1=17^3=4913 \implies 16p=4912 \implies p=307[/math], the only prime number [math]p[/math] such that [math]16p+1[/math] is the cube of some positive integer.∎
Prove: n^5-n is divisible by 10 for all n>=2?
Base Case: Show that the claim is true for n = 2. 2^5 - 2 = 30 = 3 * 10. Inductive Step: Assuming that the claim is true for n = k: (k + 1)^5 - (k + 1) = (k + 1) [(k + 1)^4 - 1] = (k + 1) [(k^4 + 4k^3 + 6k^2 + 4k + 1) - 1] = (k + 1) (k^4 + 4k^3 + 6k^2 + 4k) = k^5 + 5k^4 + 10k^3 + 10k^2 + 4k = (k^5 - k) + 5(k^4 + k) + 10(k^3 + k^2) By the inductive hypothesis, we know that k^5 - k is divisible by 10. Clearly, 10(k^3 + k^2) is divisible by 10. Finally, 5(k^4 + k) is divisible by 10. (For this last fact, it suffices to show that k^4 + k is even. This is easily shown by cases: If k is even/odd, then k^4 is also even/odd as well. Hence, k^4 + k is even.) Therefore, (k + 1)^5 - (k + 1) is also divisible by 10, showing that the claim is also true for n = k+1. ------------------- Hence, the claim is true for all integers n ≥ 2 by the Principle of Mathematical Induction. I hope this helps!
What is the result of studies on children and meditation?
I cannot speak to study data but I can tell you what I have seen in my 9 years as a meditation teacher: children who are started on a formal meditation practice very early frequently have problems, sometimes very serious, for the remainder of their lives.I am not sure as to cause and effect - if it is the meditation or some other factor - but I have seen it with almost 100% regularity in those I have met who began at an early age and are now adults. Almost none of them attribute their problems to meditation believing their early work made them somehow cool.I do not teach my five year old daughter to meditate but I do work with her on concentration skills in a very light and easy way. My goal is to set the frame for a later practice