30xx-28x+6=0
1. Take out common factor (2):
2(15xx-14x+3)=0
2. Multiply first and third terms:
15*3=45
3. Factor 45, find factors that add up to the second term (-14):
45=3*3*5
45=9*5, -5*-9
45=15*3, -15*-3
4. Rewrite equation:
2(15xx-14x+3)=0
2(15xx-5x-9x+3)=0
5. Factor pairs of terms (ie first and second, third and fourth):
2(15xx-5x-9x+3)=0
2(5x(3x-1)-3(3x-1))=0
6. Take out common factor:
2(3x-1)(5x-3)=0
Rule: If the product of a set of terms is zero, one or more of the terms must be zero
2 /= 0
3x-1=0 --> 3x=1 --> 1/3=x
5x-3=0 --> 5x=3 --> 3/5=x
7 and 9 are factors and divisors of 63 63 is a multiple and a product of 7 and 9 63 is divisible by 7 and 9
252 2x126 2x2x63 2x2x3x3x7
2^2 x 3 x 5
True yal :)
As a product of its prime factors: 2*2*2*3*5 = 120
7 and 9 are factors and divisors of 63 63 is a multiple and a product of 7 and 9 63 is divisible by 7 and 9
24 = 23 x 3
4 is a factor of 28. 4 is a divisor of 28. 7 is a factor of 28. 7 is a divisor of 28. 28 is a multiple of 7. 28 is a multiple of 4. 28 is a product of 4 and 7. 28 is divisible by 4. 28 is divisible by 7.
252 2x126 2x2x63 2x2x3x3x7
As a product of its prime factors: 7*7*13 = 637
2^2 x 3 x 5
Using the digits, we can make 81 x 62 equals 5022, which is the largest possible product.
As a product of its prime factors in exponents: 22*3*7 = 84
True yal :)
Using the quadratic equation formula the solutions are:- x = -27.0382854 and x = 90.5382854
You can not use the number 1, 2, 3, 4 to get a product 100. This is because 100 also has 5 as its factor. So you cannot get the product 100 without using 5.
45 x 68 = 3060