Factoring:
ax**2 + bx + c
Step 1: Multiply coefficients of first and third terms
a∗c
1*54
Step 2: Factor product of coefficient of first and third term (a*c), choose 2
factors whose sum is the coefficient of the second term (b)
a∗c=d∗e
d+e=b
54=2*3*3*3
54=2*27
54=6*9
54=18*3
9+6=15
Step 3: Rewrite equation
ax**2 + dx + ex + c
x**2+ 9x+6x+54
Step 4: Factor sets of terms (first and second, third and fourth)
gx(hx + i) + j(hx + i)
x(x+9)+6(x+9)
where: g is the greatest common factor of a and d
where: j is the greatest common factor of e and c
where: h = a ÷ g, e ÷ j
where: i = d ÷ g, c ÷ j
Step 5: Combine Like terms:
(gx + j)(hx + 1)
(x+6)(x+9)
This question does not make sense. If it should be X2 x 15X -54, the answer is: X2 -15X + 54 = (X - 6)(X -9)
x2 + 15x + 54 = (x + 6)(x + 9) or (x + 9)(x + 6) since 6*9 = 54 and 6 + 9 = 15However, commutativity is implicit in deriving this factorisation so that the two factorisations are the same.or factor by groupingx2 + 15x + 54= x2 + 6x + 9x + 54= (x2 + 6x) + (9x + 54)= x(x + 6) + 9(x + 6)= (x + 6)(x + 9)
The factors of 50 are 1,2,5,10,25, and 50. The factors of 54 are 1, 2, 3,6,9,18,27, and 54. Are you happy!
Factors of X2 - 16 are (X + 4) and (X - 4).
The factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54. Of these, 2 and 3 are prime.
This question does not make sense. If it should be X2 x 15X -54, the answer is: X2 -15X + 54 = (X - 6)(X -9)
x2 + 15x + 54 = (x + 6)(x + 9) or (x + 9)(x + 6) since 6*9 = 54 and 6 + 9 = 15However, commutativity is implicit in deriving this factorisation so that the two factorisations are the same.or factor by groupingx2 + 15x + 54= x2 + 6x + 9x + 54= (x2 + 6x) + (9x + 54)= x(x + 6) + 9(x + 6)= (x + 6)(x + 9)
The correct classification of x2 15x is a monomial.
x2-15x+54 factor pairs of 54: 1,54 2,27 3,18 6,9 -1,-54 -2,-27 -3,-18 -6,-9 <-6 and -9 sum to -15 x2-6x - 9x+54 x(x-6) - 9(x-6) (x-6)(x-9)
3x2 + 15x + 6 = 3*(x2 + 5x + 2). There are no further rational factors.
(x-6)(x-9) I did that by factoring 54 until I found two numbers that added up to 15.
x2+15x+14 = (x+1)(x+14)
(x - 6)(x - 9)
You're looking for factors of 100 that differ by 20. (x + 5)(x - 20)
You can't factor it
It's an equation. If you want to solve it for x, you can do so as follows: x2 + 15x = 30 x2 + 15x + (15/2)2 = 30 + (15/2)2 (x + 15/2)2 = 30 + 225/2 x + 15/2 = ± √(285/2) x = (-15 ± √570) / 2
(x + 7)(x + 8)