Answer: I hope this is the correct answer for you.
Step-by-step explanation:
2x^2 + x - 6 = 2x^2 + 4x - 3x - 6
= 2x(x + 2) - 3(x + 2) = (x +2)(2x - 3) (Ans)
Or by completing the square method,
2x^2 + x - 6 = 2[x^2 + (1/2)x] - 6
=2[x^2 + (1/2)x + (1/4)^2 - (1/4)^2] - 6
=2[(x+1/4)^2 - 1/16] - 6
=2(x + 1/4)^2 - 1/8 - 6
=2(x + 1/4)^2 - 49/8
=2[(x + 1/4)^2 - 49/16]
=2[(x + 1/4)^2 - (7/4)^2]
=2[x + 1/4 + 7/4][x + 1/4 -7/4]
=2(x + 2)(x - 3/2)
=(x + 2)(2x - 3) (Ans)
f(x) = a x2 + b x + c
then
f(x) = a (x - x1) (x - x2)
For your problem we have the roots given by
\displaystyle \frac{-1 \pm \sqrt{1^2 - 4 (2) (-6)}}{2 (2)}= \frac{-1 \pm \sqrt{49}}{4} = \frac{-1 \pm 7}{4}
2(2)
−1±
1
2
−4(2)(−6)
=
4
−1±
49
=
4
−1±7
which gives -2 and 3/2 as the roots
So
f(x) = 2x2 + x - 6 = 2 (x + 2) (x - 3/2) = (x + 2) (2 x - 3)
Which, thankfully, is the same thing given above.