Question

2x3-5x2-28x+15 (x-5)​

Answer 1

SOLUTION

IF X-5 IS THE ROOT OF THE EQ WHICH MEANS

X-5 = 0

X= 5

PUTTING X = 5 IN EQ 2x³-5x²-28x+15

= 2(5)³-5(5)²-28(5)+15

250-125-140+15

125+25-140

= 150-140 = 10

Hence the remainder is not 0 x-5 is not root of the this eq

Answer 2

1. body mass method
2. 2×3_5×2_28+15(×_5)
3. 6_10_28×+15
4. _91+13=_78