Factor 3x^2-11x+10 and 2t^3-8t?
well 6+5 = 11 so work with 2*3 and 5
(3x - 5)(x - 2)
2t^3-8t
take 2 t out first
2t(t^2 - 4)
now you have the difference of two squares on the right
2 t (t-2)(t+2)
thank you very much (:
To factor the expression 3x^2 - 11x + 10, we need to find two binomials whose product gives us this expression. We can use the factoring method known as "AC Method" or "Splitting the Middle Term."
The general form of a quadratic expression is ax^2 + bx + c. In this case, a = 3, b = -11, and c = 10. We need to find two numbers whose product is ac (3*10 = 30) and whose sum is b (-11).
The numbers that satisfy this condition are -6 and -5 because (-6) * (-5) = 30 and (-6) + (-5) = -11.
Next, we rewrite the middle term (-11x) as the sum of these two terms:
3x^2 - 6x - 5x + 10.
Now, we can factor by grouping:
(3x^2 - 6x) + (-5x + 10).
From the first group, we can factor out 3x:
3x(x - 2).
From the second group, we can factor out -5:
-5(x - 2).
Combining these factors, we have:
3x(x - 2) - 5(x - 2).
Notice that we now have a common factor of (x - 2). We can factor it out:
(x - 2)(3x - 5).
Therefore, the factored form of 3x^2 - 11x + 10 is (x - 2)(3x - 5).
To factor the expression 2t^3 - 8t, we can first factor out the greatest common factor (GCF), which, in this case, is 2t:
2t(t^2 - 4).
Now we have the expression 2t multiplied by the difference of squares, t^2 - 4. The difference of squares can be further factored as follows:
2t(t + 2)(t - 2).
Therefore, the factored form of 2t^3 - 8t is 2t(t + 2)(t - 2).