ib hl adrian sparrow factor and remainder theorem
TRANSCRIPT
Factors and solutions
Factorise the quadratic x2-x-12. (x+3)(x-4)
Solve x2-x-12=0 x=-3 and x=4
Substitute x=-3 into the equation, and calculate the answer.
Repeat with x=4.
Both answers are 0, because these are factors.
Cubic polynomials
This technique will also work for higher order polynomials such as cubics.
Show that (x-3) is a factor x3-4x2+x+6.
Substitute in x=3. (3 3)-4(32)+3+6=0
Hence, using trial and error find the other 2 factors of x3-4x2+x+6.
Hint: these factors are usually integers, start with 0, then try 1, -1, 2, etc.
(x-2) (x+1)
Cubic polynomials - typical questions
Given that (x-2) and (x+1) are factors of the cubic polynomial 2x3+ax2+bx+6, find the values of a and b.
Substitute in the first factor x=2, make up a linear equation.
Substitute in the second factor x=(-1), make up another linear equation.
Solve simultaneously to find a and b.
2(23)+(22)a+2b+6=0
2(-13)+(-12)a-b+6=0
16+4a+2b+6=0
4a+2b=-22
-2+a-b+6=0
a-b=-4
a=-5, b=-1
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(x + 3) x 2 − x −12)
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x
Dividing by factors - quadratics
This technique is not used by the IB, but is useful to know and explains the remainders - division by an algebraic factor.
Divide the quadratic x2-x-12 by one of it’s factors, e.g. (x+3).
Divide x2 by x.
Take the answer and multiply by 3.
Take this away from the next component in the polynomial.
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−3x
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−4xDivide this answer by x.
Repeat the steps above.
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−4
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−12 -
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−12
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0Two things to note:
Remainder = 0The other factor is given as the answer.
Now repeat this process but divide the quadratic by (x-4)
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(x − 3) x 3 − 4x 2 + x + 6)
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−3x 2
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+3x
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−2x + 6
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+ 6
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0€
−x 2 + x
Dividing by factors - cubics
Divide the polynomial x3-4x2+x+6 by (x-3).
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x 2 − x −2
The quotient left as an answer can now be factorised to give the other two factors of the original cubic.
All 3 factors are:
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(x −2)(x +1)(x − 3)€
(x −2)(x +1)
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x 2 − x −2
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(x −2) x 3 − x 2 −10x −8)
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x 2 + x −8
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−2x 2
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x 2 −10x
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−2x
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−8x −8
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+16
Remainders when dividing
Divide the polynomial x3-x2-10x-8 by (x-2).
We have a remainder.
This tells us that (x-2) is not a factor of the polynomial.
Substitute x=2 into the polynomial x3-x2-10x-8.
Answer = -24€
−24
Using the remainder
Given that (x+1) is one of the factors of the cubic polynomial ax3+5x2+bx-9, and that when the polynomial is divided by (x-2) a remainder of 15 is left, find the values of a and b.
Substitute in the first factor x=-1, make up a linear equation.
Substitute in the value x=2, but the answer is 15, not 0. Make up a linear equation.
Solve simultaneously to find a and b.
a(-13)+5(-12)-b-9=0
a(23)+5(22)+2b-9=15
-a+5-b-9=0
a+b=-4
8a+20+2b-9=15
8a+2b=4
a=2, b=-6
Exam questions
2. Given that (x+1) is one of the factors of the cubic polynomial 3x3+ax2+bx+4, and that when the polynomial is divided by (x-1) a remainder of -12 is left, find the values of a and b.
1. Given that (x-2) and (x-5) are both factors of the cubic polynomial ax3+bx2-x+30, find the values of a and b.
3. a) Show that (x-3) is one of the factors of the cubic polynomial 2x3-3x2-14x+15.
a=2, b=-11
a=-10, b=-9
b) Hence find the remaining two factors of the cubic polynomial 2x3-3x2-14x+15.
2(33)-3(32)-14(3)+15=0
Hint: you will need to divide by (x-3).
(x-3)(x-1)(2x+5)