B. (x³ + 6x² - x-6) ÷ (x + 6)
Answers
Answer:
If you assume that at least one of the roots is a whole number, than we can say that at least one of the roots is either ±1±1 , ±2±2 , ±3±3 , ±6±6
Why?
If you consider the following polynomial of third degree.
(x±a)(x±b)(x±c)(x±a)(x±b)(x±c)
(Think of this as the factorised form of the equation given).
⟹(x2±bx±ax±ab)(x±c)⟹(x2±bx±ax±ab)(x±c)
=(x3±x2c±bx2±bxc±ax2±axc±abx±abc)=(x3±x2c±bx2±bxc±ax2±axc±abx±abc)
(Think of this as the expanded form of the equation given).
Here we have that given this information, than assuming that the equation factorises semi neatly with at least one root as a whole number, than we can test for the roots that are factors of -6.
If the possible roots include ±1±2±3±6±1±2±3±6 , than our possible factors are (x±1),(x±2),(x±3),(x±6)(x±1),(x±2),(x±3),(x±6) .
We can test if just one of these roots is true by polynomial division (as we are then left with a quadratic which we can factorise - regardless if it doesn’t factorise neatly or not).
Performing polynomial division reveals that (x−1)(x−1) is a factor.
We have from this that.
x3−6x2+11x−6x−1=x2−5x+6x3−6x2+11x−6x−1=x2−5x+6
Concluding from this, we know the following is true.
(x−1)(x2−5x+6)=0(x−1)(x2−5x+6)=0
The second polynomial factorises into.
(x2−5x+6)=(x−3)(x−2)(x2−5x+6)=(x−3)(x−2)
Thus we have the final factorised form where.
(x−1)(x−3)(x−2)=0(x−1)(x−3)(x−2)=0
So we have that.
x=1,2,3x=1,2,3
And just to check we can expand our factorised form which gives.
(x2−4x+3)(x−2)=0(x2−4x+3)(x−2)=0
x3−6x2+11x−6=0x3−6x2+11x−6=0
This checks out and if you’re unsure about polynomial long division I would recommend watching Eddie Woo on
here is an example for you...
hopes it helps you...
x³ + 6x² - x - 6 / x + 6
= x³ + 6x² - x + 6 / x + 6
= x² +6x² -x ans
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