### Food for Thought

SmartAlec says:
When I expand (2x + 7)(x - 1), I get 2x^2 + 5x - 7.
I can use my calculator to help me check if my answer is correct.

For instance, if x = 2,
(2x + 7)(x - 1) = 11
and
2x^2 + 5x - 7 = 11
That means I have done my expansion correctly.

I can safely say, I can do this with any value of x and for any given expression.

Do you agree? Justify your argument with example(s).

1. No. If x is 0 and if is something/x it would be undefined.

2. No. if x=3
2(x)^2+5x-7=2(3)^2+5(3)-7=26

3. No. x may be 0 (highly unlikely) and then it will be wrong

4. I do not agree. It works almost all the time, but if the x is 0, the equation will not work.

5. It does not apply to all numbers.

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7. no if x is 0 or (-2) then it will wrong

8. Dose not works with all the number

9. No. If x = 0 or x = 1/3 or other fractions that cannot be converted into a decimal.

10. No. There is a chance it might be zero, then the answer will be different.

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13. No, I would not work all the time.
e.g. Let x be 5.
(2x + 7)(x - 1)=68
2x^2 + 5x - 7=118

14. If the value of X increases, the value of the equation gets higher. Since ^2 is exponential, that means that 2^2=4, but if you use a bigger number like 3^2=9, it is different.

Example:
Assume that X is 4,
2X4^2 + 5X4 - 7 = 45, not 11.

15. No, I do not agree as x can also be a negative number and
For instance, if x is -2
(2x+7)(x-1)=(-4+7)(-2-1)=3*-3=-9
and
2x^2 + 5x - 7 = 16-10 + -7 = -1

16. It does not apply all the time when x=3

17. I do not agree as all the other values cannot work.

18. NO. Example, if the question is x/x-5 - 3/x+7, and the value of x is 5, the denominator would be 0, and the calculator would have an error.

(stated by ms loh xD)

19. Let x = 11

2(x)^2+5(x)-7

= 2(11)^2+5(11)-7
= 22^2+55-7
= 484+55-7
= 534

:O

20. It works with all whole numbers except for 5 and 0.