Synthetic Division! Viable roots are (multiples of a)/(multiples of d) so half, -half of, 2, -2, 1, -1 [a is 2 and d is -4, they are the first and last numbers] Then simply start going by way of the viable roots in synthetic division. 1 2 7 -5 -four 2 9 4 zero One worked, so now we've (x-1)(2x^2 +9x +4) -half of 2 9 4 -1 -four 0 -4 -1 -4 four zero terrible one 1/2 worked, and so did poor four! So now we've (x-1)(2x+1)(x+four) Roots are 1, -half of, and -four!
Answers & Comments
7x – 2(x-2) + 5(x+3)
= 7x -2x +4 +5x+15
= 5x + 4 + 5x + 15
= 10x + 19
7x - 2(x - 2) + 5(x + 3) =
7x - 2x + 4 + 5x + 15 =
10x + 19.
7x - 2(x - 2) + 5(x + 3)
7x - 2x + 4 + 5x + 15
10x + 19
7x - 2 (x - 2) + 5 (x + 3)
= 7x - 2x + 4 + 5x + 15 expand what is in the bracket and write it out
= 7x - 2x + 5x + 4 + 15 group all the x-es and numbers together
= 10x + 19 subtract/add the x-es and the numbers
7x-2x+4+5x+15= 10x+19
Expand out each set of brackets:
2(x-2) = (2x - 4)
5(x+3) = (5x + 15)
Put it all back together...
7x - (2x - 4) + (5x + 15) ---> (remove brackets) 7x - 2x + 4 + 5x + 15 ---> (simplify) 10x + 19
10x + 19
Synthetic Division! Viable roots are (multiples of a)/(multiples of d) so half, -half of, 2, -2, 1, -1 [a is 2 and d is -4, they are the first and last numbers] Then simply start going by way of the viable roots in synthetic division. 1 2 7 -5 -four 2 9 4 zero One worked, so now we've (x-1)(2x^2 +9x +4) -half of 2 9 4 -1 -four 0 -4 -1 -4 four zero terrible one 1/2 worked, and so did poor four! So now we've (x-1)(2x+1)(x+four) Roots are 1, -half of, and -four!