x3 + 2x2 - 19x - 20 = x3 + x2 + x2 + x - 20x - 20 = x2(x + 1) - x(x + 1) - 20(x + 1)
= (x + 1)(x2 + x - 20) = (x + 1)(x - 4)(x + 5)
a2+9a+20 = (a+4)(a+5) when factored
m2 + 9m + 20 = (m + 5) (m + 4)
20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 = 400.
exactly 20 of them
2*2 - 12x +16 =0 4 - 12x +16 =0 -12x + 16 +4 =0 -12x + 20 =0 -12x = -20 x = 1.666666667
19x+1x+37=0 First add the 19x and 1x together 20x+37=0 Next subtract 37 to the other side 20x=-37 Now divide both sides by 20 x=-20/37
4x2+2x-20 2x2+x-10 2x2+5x-4x-10 x(2x+5)-2(2x+5) (x-2)(2x+5)
That doesn't factor neatly. Applying the quadratic formula, we find two real solutions: 10 plus or minus 4 times the square root of 7. x = 20.583005244258363 x = -0.5830052442583629
2x2 + 3x - 20 = (x + 4)(2x - 5).
2x2+18x = 20 2x2+18x-20 = 0 x2+9x-10 = 0 (x+10)(x-1) = 0 x = -10 or x = 1
16=2x2x2x220=2x2x5GCF=2x2=4
If you are talking of 20 and 24 then factor both to prime factors. 20=2x2x5 24=2x2x2x3 circle the common ones. 2x2=4 the greatest common factor.
how do you factor y=2x^2+3x-20 to get the roots
2 x 2
a2+9a+20 = (a+4)(a+5) when factored
We figure this out by dividing 500 by 25. The answer is 20, so the ratio is 1 teacher for 20 students.
-4x2 + 6x + 40 = 0 ∴ -2x2 + 3x + 20 = 0 ∴ -2x2 - 8x + 5x + 20 = 0 ∴ -2x(x + 4) + 5(x + 4) = 0 ∴ (5 - 2x)(x + 4) = 0 ∴ x ∈ {5/2, -4}