The correct option is A 2(x+2)(x−3)(x−4)
p(x)=(2x3−10x2−4x+48)
Using trial and error method,
when
x=−2p(−2)=2×(−2)3−10×(−2)2−4×(−2)+24) =−16+40−8−48=0
Hence, one factor = (x+2)
Using synthetic division,
x3x2x1x02−10−448−2↓−428−482−1424|−0
∴p(x)=(x−2)(2x2−14x+24)
Now, using trial and error method,
when,
x=3q(x)=(2x2−14x+24) =2×32−14×3+24=0
Hence, another factor is (x−3)
x2x1x02−14243↓6−242−8|−0
∴p(x)=(x+2)(x−3)(2x−8) =2(x+2)(x−3)(x−4)