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How would you prove that x plus 2 is a factor of the expression 2x cubed -3x squared -12x plus 4?
Wiki User
∙ 12y ago
By (long) division:
. . . . . . . . . . .2x2 - 7x . + 2
. . . . . ----------------------
x + 2 | 2x3 - 3x2 - 12x + 4
. . . . . .2x3 + 4x2
. . . . . .-----------
. . . . . . . . . - 7x2 - 12x
. . . . . . . . . - 7x2 - 14x
. . . . . . . . . ------------
. . . . . . . . . . . . . . . . 2x + 4
. . . . . . . . . . . . . . . . 2x + 4
. . . . . . . . . . . . . . . . -------
. . . . . . . . . . . . . . . . . . . . .0
. . . . . . . . . . . . . . . . ====
(the "dot-spaces" are used to hold the characters in the right place of the division - they should be treated as blank)
Thus since
(x + 2)(2x2 - 7x + 2) = 2x3 - 3x2 - 12x + 4
(x + 2) is a factor of 2x3 - 3x2 - 12x + 4
In the division:
- the first term of the divisor (x) is compared with the highest power of x remaining in the dividend to find the next term of the quotient;
- the whole divisor is multiplied by this;
- then subtracted from the dividend;
- Steps 1-3 are repeated until there is no first term of the divisor (x) in the dividend.
- If the dividend is '0' (ie the last multiplication resulted in what was remaining in the dividend) then the divisor is a factor of the original dividend); otherwise it is not a factor.
Wiki User
∙ 12y ago
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