DO NOT REMOVE FROM CLASS.
QUICK CHECK 3A - REASONING KEY: USE RED PEN
Recall and Reproduction:
1. (LT 2A) Divide using Algebraic Long Division
(3 + 8𝑦 3 − 6𝑦 2 − 7𝑦) ÷ (2𝑦 − 3)
First we need to set the dividend into standard form.
2𝑦 − 3|8𝑦 3 − 6𝑦 2 − 7𝑦 + 3
We then can go through the process of dividing the first term of the dividend by the first term of the
divisor, multiplying that result by the divisor, and finally subtracting the result with the dividend.
4𝑦 2
Divide the 8y3 with 2y to get 4y2.
Multiply
4y2
by 2y – 3 to get 8𝑦 − 12𝑦
3
2
Distribute the negative.
Combine like terms.
Bring down the next term.
Repeat the steps. The remainder will be 6 divided
by the dividend
2𝑦 − 3|8𝑦 3 − 6𝑦 2 − 7𝑦 + 3
4𝑦 2
2𝑦 − 3|8𝑦 3 − 6𝑦 2 − 7𝑦 + 3
8𝑦 3 − 12𝑦 2
4𝑦 2
2𝑦 − 3|8𝑦 3 − 6𝑦 2 − 7𝑦 + 3
−(8𝑦 3 − 12𝑦 2 )
4𝑦 2
2𝑦 − 3|8𝑦 3 − 6𝑦 2 − 7𝑦 + 3
−8𝑦 3 + 12𝑦 2
4𝑦 2
2𝑦 − 3|8𝑦 3 − 6𝑦 2 − 7𝑦 + 3
−8𝑦 3 + 12𝑦 2
0 + 6𝑦 2 − 7𝑦
𝟒𝒚𝟐 + 𝟑𝒚 + 𝟏 +
𝟔
𝟐𝒚−𝟑
2𝑦 − 3|8𝑦 3 − 6𝑦 2 − 7𝑦 + 3
−8𝑦 3 + 12𝑦 2
+6𝑦 2 − 7𝑦
−(+6𝑦 2 − 9𝑦)
2𝑦 + 3
−(2𝑦 − 3)
6
1)
𝑃(𝑥)
𝑅
(LT 1C) Use synthetic division to write the quotient 𝑃(𝑥) ÷ (𝑥 − 𝑟) in the form
= 𝑄(𝑥) +
where R is a
𝑥−𝑟
𝑥−𝑟
constant.
(3𝑥 3 − 4𝑥 2 − 7𝑥 + 9) ÷ (𝑥 − 2)
To do synthetic division, first put all of the coefficients of the dividend across the top.
We use the opposite of r because the root of x-2 is 2.
We than do the algorithm of dropping the first term, multiplying by 2, putting what we get under the
second term. We add the second term and what we got to get the next number. Repeat.
2
3
-4
-7
9
6
4
-6
3
2
-3
3
The numbers at the bottom of the synthetic division table represents the coefficients of the quotient. The
last term will be the remainder. The numbers before it go up in degree. -3 is the constant, 2 is the xterm, and 3 is the x2 term.
3𝑥 3 − 4𝑥 2 − 7𝑥 + 9
𝟑
= 𝟑𝒙𝟐 + 𝟐𝒙 − 𝟑 +
𝑥−2
𝒙−𝟐
Routine:
2) (LT 2C) Find all roots exactly for the polynomial equation
𝑥 4 + 2𝑥 3 − 2𝑥 2 − 6𝑥 − 3 = 0
To find all the roots, we need to first use the rational zero theorem to find how what possible rational
roots there are. We know there will be up to 4 roots because of the Fundamental Theorem of Algebra.
Rational Zero Theorem:
𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑝 ±1, ±3
=
= ±1, ±3
𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑞
±1
Therefore, 1, -1, 3, and -3 are the possible rational zeroes.
We can now test these by using the remainder theorem and synthetic division.
1
2
-2
-6
1
1
3
1
-5
-1
1
1
-3
-3
3
1
5
13
33
-3
1
-1
1
-9
We now know that there is an x-intercept or zero at -1.
The coefficients of the new equation will give us two factors. Note the factor of the zero.
𝑥 4 + 2𝑥 3 − 2𝑥 2 − 6𝑥 − 3 = (𝑥 3 + 𝑥 2 − 3𝑥 − 3)(𝑥 + 1)
We can now do the same process for the new 3rd degree polynomial.
Rational Zero Theorem:
𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑝 ±1, ±3
=
= ±1, ±3
𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑞
±1
Therefore, 1, -1, 3, and -3 are the possible rational zeroes.
We can now test these by using the remainder theorem and synthetic division.
1
1
-3
-3
1
1
2
-1
-4
-1
1
0
-3
0
3
1
4
9
24
-3
1
-2
3
-12
There is another 0 at -1 meaning that it is a double root.
The coefficients of the new equation will give us two factors. Note the factor of the zero.
𝑥 4 + 2𝑥 3 − 2𝑥 2 − 6𝑥 − 3
(𝑥 3 + 𝑥 2 − 3𝑥 − 3)(𝑥 + 1)
(𝑥 2 − 3)(𝑥 + 1)(𝑥 + 1)
Finally we have a quadratic function which lets us solve for x using traditional methods.
In this case, we can just use algebra because there is no b term.
𝑥2 − 3 = 0
𝑥2 = 3
√𝑥 2 = ±√3
𝑥 = ±√3
Therefore, the roots of the function 𝑥 4 + 2𝑥 3 − 2𝑥 2 − 6𝑥 − 3, are: 𝑥 = −1, −1, √3, −√3
-3
-8
0
96
24