Factor
18-9t+t^2
I don't think this is factorable help
how about
(6-t)(3-t) ?
oh yes the 18w^2 is posative Thank you
To determine if the quadratic expression 18-9t+t^2 is factorable, we can use the quadratic equation of the form ax^2+bx+c.
In this case, a = 1, b = -9, and c = 18. We can determine if the expression is factorable by calculating the discriminant, which is given by the formula discriminant = b^2 - 4ac.
Substituting the values into the formula, we get:
discriminant = (-9)^2 - 4(1)(18) = 81 - 72 = 9
Since the discriminant is positive (greater than 0), the quadratic expression is factorable.
To factor the expression, we can use the fact that the quadratic expression can be written as the product of two binomials. The form of these binomials is: (tx + a)(cx + b), where a, b, c, and d are constants.
To find the values of a, b, c, and d, we need to look at the coefficients in front of t in the original expression 18-9t+t^2.
The coefficient of t^2 is 1, so we have (tx + a)(cx + b) = t^2x^2 + ...
The coefficient of t is -9, so we have (tx + a)(cx + b) = t^2x^2 - (ac)x - (bc) = 1x^2 - (ac)x - (bc) ...
The constant term is 18, so we have (tx + a)(cx + b) = t^2x^2 - (ac)x + 18 = 1x^2 - (ac)x - (bc) + 18...
Now, we need to find two numbers that multiply to give a product of 18 and add up to -9. The numbers that satisfy this condition are -6 and -3.
Thus, we have:
(1x + (-6))(1x + (-3)) = (x - 6)(x - 3)
Therefore, the factored form of the quadratic expression 18-9t+t^2 is (t - 6)(t - 3).
I hope this explanation helps you understand how to factor the expression.