x2-18x+81 = (x-9)(x-9) when factored
Divide all terms by 3 and so x2+18x+81= (x+9)(x+9) when factored
To factor this, find two numbers whose product is 81 and sum 18. These two numbers are therefore 9 and 9. So we can rewrite the polynomial in its factored form as follows: (k+9)(k+9), or "k plus 9 squared"
78 + 79 + 80 + 80 + 81 + 81 = 479
30a2+89a+63 = (5a+9)(6a+7) when factored
There's no root and there's no variable, so no, it can't be factored, but it can be simplified to 113.
It is 8(80b+81) when factored
x2-18x+81 = (x-9)(x-9) when factored
Divide all terms by 3 and so x2+18x+81= (x+9)(x+9) when factored
9x2 - 6xy + y2 - 81 = (3x - y)2 - 92 = (3x - y - 9)(3x - y + 9)
To factor this, find two numbers whose product is 81 and sum 18. These two numbers are therefore 9 and 9. So we can rewrite the polynomial in its factored form as follows: (k+9)(k+9), or "k plus 9 squared"
It is (x+4)(x+5) when factored
78 + 79 + 80 + 80 + 81 + 81 = 479
2(a+b) is 2a plus 2b in factored form.
30a2+89a+63 = (5a+9)(6a+7) when factored
w2+18w+77 = (w+7)(w+11) when factored
2x2+5x+3 = (2x+3)(x+1) when factored