q3-q2+2q-2 = (q-1)(q2+2) = (q-1)(q+2.5i)(q-2.5i)
(a + 2b)(a + 2b)
Since the problem has 4 terms, first you factor x cubed plus 9x squared, then you factor 2x plus 18. So when you factor the first two term, you would get x sqaured (x plus 9). Then when you factor the last two terms and you get 2 (x plus 9). Ypure final answer would be (x squared plus 2)(x plus 9)
Two is a prime factor of that equation.
The only factor is 2. 2*(t3 + 2t2 + 4x)
You can factor a five out of 15 plus 5x. Therefore, 15 plus 5x factors out to 5(3+x). Hope it helps :)
5m
The other factor is 1.
(a + 2b)(a + 2b)
Since the problem has 4 terms, first you factor x cubed plus 9x squared, then you factor 2x plus 18. So when you factor the first two term, you would get x sqaured (x plus 9). Then when you factor the last two terms and you get 2 (x plus 9). Ypure final answer would be (x squared plus 2)(x plus 9)
factor the trinomial 16x^2+24x+9
You can't factor it
That does not factor neatly.
Factor x2 plus 12xp plus 36p2 is (x+6p)(x+6p).
No. √18 cannot be expressed as a fraction of the form p/q. 18 = 2 x 9 = 2 x 32 ⇒ √18 = √(2 x 32) = (√2) x 3. So if √2 is rational then √18 is rational. Assume √2 is rational. Then p and q can be found such that √2 = p/q is in its simplest form, that is p and q have no common factor. Consider: (√2)2 = (p/q)2 ⇒ 2 = p2/q2 ⇒ p2 = 2q2 Thus p2 is even, and so p must be even. Let p = 2r. Then: p2 = (2r)2 = 2q2 ⇒ 4r2 = 2q2 ⇒ 2r2 = q2 Thus q2 is even, and so q must be even. Let q = 2s. Thus p = 2r, q = 2s and so p and q have a common factor of 2. But p and q are such that they have no common factor. Contradiction. Thus the assumption that √2 is rational is false, that is √2 is not rational, so √18 is not rational.
1
No
Not factorable