Q: What is the answer to P not a factor of 6?

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m(6 + p)

x-a is a factor of the polynomial p(x),if p(a)=0.also,if x-a is a factor of p(x), p(a)=0.

x2 + 13x + 4 = (x + 6½ + √38¼)(x + 6½ - √38¼). To find this, we need to find p and q, where p + q = 13, pq = 4. 4 = 42¼ - 38¼ = (6½ + √38¼)(6½ - √38¼); thus, p = (6½ + √38¼), q = (6½ - √38¼).

6p - pq 7pr factorizing = -1

divide the fraction by a common factor. Here p, it is 2. 6/2 is 3 and 10/2 is 5 so it is 2/5

Related questions

6(p - 18)

m(6 + p)

You have to use the quadratic formula: p=(6±√(36+72))/2 p=(6±√108)/2 p=3±3√3

x-a is a factor of the polynomial p(x),if p(a)=0.also,if x-a is a factor of p(x), p(a)=0.

Yes it is. The proof is as follows:We prove the statement by contradiction i.e. Assume that sqrt(6) is a rational number.Then there exist positive integers p and q with gcd(p,q) = 1 such that p/q = sqrt(6).Square both sides: p^2 / q^2 = 6,p^2 = 6q^2.Now as 2 is a divisor of the right-hand side (RHS), it implies that 2 is also a divisor of the left-hand side (LHS).This is only possible if 2 is a factor of p.Let p =2k. Then k is a positive integer as well.Thus, 4k^2 = 6q^2,2k^2 = 3q^2.As 2 is a factor of the LHS, 2 is also a factor of the RHS.But this is only possible if 2 is a factor of q.=> gcd(p,q) >= 2. Contradiction!Thus sqrt(6) is irrational.yes it is

x2 + 13x + 4 = (x + 6½ + √38¼)(x + 6½ - √38¼). To find this, we need to find p and q, where p + q = 13, pq = 4. 4 = 42¼ - 38¼ = (6½ + √38¼)(6½ - √38¼); thus, p = (6½ + √38¼), q = (6½ - √38¼).

Suppose p(x) is a polynomial in x. Then p(a) = 0 if and only if (x-a) is a factor of p(x).

6p - pq 7pr factorizing = -1

p2 + 9p + 18/ p + 6(p + 6)(p + 3)/ p + 6(p + 6)(p + 3)/ p + 6p + 3

No, 18 is a multiple of 6. 6 is a factor of 18.

6 is a factor of 54.

p/6