NOT CALCULUS. Use long division. 2x2 + x - 1(4x^4) 2x^2 goes into 4x^4 2x^2 times. The remainder will then be (2x^2*(x-1)). This result(2x^3 - 2x^2) is what need to be subtracted from 4x^4 to make it exactly divisible
16x2-9
it would simply be written as follows: 4x-9
18x + 5
The math problem 4x plus 7 times 31 equals 8. This is taught in math.
(4x-20) = 119
NOT CALCULUS. Use long division. 2x2 + x - 1(4x^4) 2x^2 goes into 4x^4 2x^2 times. The remainder will then be (2x^2*(x-1)). This result(2x^3 - 2x^2) is what need to be subtracted from 4x^4 to make it exactly divisible
16x2-9
it would simply be written as follows: 4x-9
16x
18x + 5
The math problem 4x plus 7 times 31 equals 8. This is taught in math.
4x+8
4x+2.5
4x + 2.5
3x - 7 + 4x = 28 7x - 7 = 28 I added 4x and 3x to get 7x 7x = 35 I subtracted 7 from both sides x = 5 I divided by 7 on both sides
x(x+4) equals x2 plus 4x