Solving Polynomial Equations By Factoring and Using Synthetic Division
Summary
TLDRThis tutorial focuses on solving polynomial equations using various methods, including factoring by grouping, substitution, and the rational zero theorem with synthetic division. The instructor demonstrates how to factor polynomials, identify common factors, and solve for real and imaginary roots. Examples include solving cubic and quartic polynomials, and the tutorial emphasizes recognizing patterns such as the difference of perfect squares. By the end, viewers will understand multiple techniques for solving polynomial equations, equipping them with practical methods for tackling different types of polynomial problems.
Takeaways
- 🔢 The tutorial focuses on solving polynomial equations using various methods.
- 📐 Factoring by grouping is a technique used when the ratios of the coefficients of the first two terms match those of the last two terms.
- 🔄 Factoring involves taking out the greatest common factor (GCF) from each group of terms.
- 🔄 After factoring, if possible, further factor the expression, such as a difference of squares, to completely factor the polynomial.
- 🔍 When factoring by grouping is not possible, consider reducing the polynomial to a quadratic form using substitution.
- 🔄 Substitute x<sup>n</sup> with a variable (like 'a') to transform the polynomial into a quadratic form that can be factored.
- 🔢 For polynomials that cannot be factored by grouping or reduced to quadratic form, use the rational zero theorem to list possible rational zeros.
- 🔄 Synthetic division is used to find the remaining zeros of a polynomial after identifying one zero.
- 🔄 Imaginary numbers can be used to factor polynomials with a sum of perfect squares.
- 📝 The tutorial provides step-by-step examples for each method, demonstrating how to solve specific polynomial equations.
Q & A
What method can be used to solve the equation x³ + 3x² - 4x - 12 = 0?
-The equation can be solved by factoring by grouping because the ratio of the first two coefficients (1 and 3) is the same as the ratio of the last two coefficients (-4 and -12).
How is the equation x³ + 3x² - 4x - 12 factored?
-First, group the terms and factor out the greatest common factors. The equation is factored as (x + 3)(x² - 4). Then, factor x² - 4 as a difference of squares to get (x + 3)(x + 2)(x - 2).
What are the solutions to the equation x³ + 3x² - 4x - 12 = 0?
-The solutions are x = -3, x = -2, and x = 2.
How can you solve the equation x⁴ - 5x² - 36 = 0?
-This equation can be solved by substitution. Let a = x², then solve the resulting quadratic equation a² - 5a - 36 = 0, which factors to (a - 9)(a + 4).
What are the real and imaginary solutions to x⁴ - 5x² - 36 = 0?
-The real solutions are x = ±3, and the imaginary solutions are x = ±2i.
How do you solve x³ - 3x² - 6x + 8 = 0 using the Rational Root Theorem?
-First, list all possible rational zeros using the factors of the constant term and the leading coefficient. Then, use synthetic division to test each possible zero. After finding one root, use synthetic division to simplify the equation and find the remaining roots.
What are the solutions to the equation x³ - 3x² - 6x + 8 = 0?
-The solutions are x = 1, x = 4, and x = -2.
How do you apply synthetic division to find the remaining zeros after identifying one zero in a cubic equation?
-After identifying one zero using the Rational Root Theorem, apply synthetic division by dividing the original polynomial by (x - root). The quotient will be a quadratic equation that can be solved using factoring or other methods to find the remaining zeros.
How can you factor the polynomial x⁴ - 7x³ + 7x² + 35x - 60?
-Start by using the Rational Root Theorem to find a root. Once a root is identified, use synthetic division to simplify the polynomial and then factor the resulting expression.
What are the solutions to the polynomial equation x⁴ - 7x³ + 7x² + 35x - 60 = 0?
-The solutions are x = 3, x = 4, x = ±√5.
Outlines
🔍 Solving Polynomial Equations with Factor by Grouping
In this paragraph, the focus is on solving a cubic polynomial equation through factor by grouping. The example given is \( x^3 + 3x^2 - 4x - 12 = 0 \), where the ratios of the coefficients suggest using grouping. By extracting the greatest common factor (GCF) in both parts of the polynomial, the expression simplifies, and further factoring leads to the equation's solutions: \( x = -3 \), \( x = -2 \), and \( x = 2 \). This method emphasizes the efficiency of factoring when the ratio of coefficients aligns.
🔧 Using Substitution to Solve Higher-Degree Polynomials
This paragraph explains how to solve a quartic polynomial, \( x^4 - 5x^2 - 36 = 0 \), by reducing it to quadratic form through substitution. By substituting \( a = x^2 \), the equation becomes a trinomial, which is then factored. After factoring and substituting back for \( x \), the real solutions (\( x = -3, 3 \)) and imaginary solutions (\( x = \pm 2i \)) are determined, emphasizing the versatility of substitution for higher-degree polynomials.
⚖️ Solving Cubic Polynomials Using Rational Zeros and Synthetic Division
Here, a cubic equation \( x^3 - 3x^2 - 6x + 8 = 0 \) is tackled. Since factor by grouping is not viable, the rational zero theorem is used to list possible rational roots, starting with \( x = 1 \), which is verified. Synthetic division simplifies the equation to a quadratic form, allowing it to be factored, revealing the three real solutions: \( x = 4, -2, 1 \). The method showcases rational zeros and synthetic division for non-factorable polynomials.
📐 Factoring a Quartic Polynomial Using Grouping and Synthetic Division
This paragraph demonstrates solving a fourth-degree polynomial \( x^4 - 7x^3 + 7x^2 + 35x - 60 = 0 \). After testing potential rational zeros using the rational zero theorem, \( x = 3 \) is found to be a solution. Synthetic division reduces the equation, and grouping is used to factor the remaining terms. The final solutions include real numbers and irrational roots (\( x = 4, \pm \sqrt{5} \)). The paragraph highlights the effectiveness of multiple factoring techniques.
Mindmap
Keywords
💡Polynomial Equations
💡Factoring by Grouping
💡Greatest Common Factor (GCF)
💡Difference of Perfect Squares
💡Quadratic Form
💡Rational Zero Theorem
💡Synthetic Division
💡Imaginary Numbers
💡Zeros of a Function
💡Trinomial
Highlights
Introduction to solving polynomial equations using factoring techniques.
Illustrating how to factor by grouping when the ratio of the first two coefficients is the same as the ratio of the last two.
Step-by-step breakdown of factoring x^3 + 3x^2 - 4x - 12 by grouping.
Explanation of the difference of perfect squares used to factor x^2 - 4 into (x + 2)(x - 2).
Introduction of a second polynomial equation reducible to a quadratic form, x^4 - 5x^2 - 36 = 0, using substitution.
Detailed breakdown of factoring a quadratic form and how substitution simplifies complex polynomials.
Introduction to imaginary numbers when factoring x^2 + 4 using complex numbers (2i and -2i).
Application of the Rational Zero Theorem to find possible rational solutions for higher-degree polynomials.
Using synthetic division to evaluate potential rational zeros and simplify the polynomial.
Demonstrating how synthetic division is applied to the equation x^3 - 3x^2 - 6x + 8 = 0 to find all real solutions.
Further example of polynomial solving using Rational Zero Theorem and synthetic division on a fourth-degree polynomial, x^4 - 7x^3 + 7x^2 + 35x - 60.
Explanation of factoring by grouping to simplify the equation into x^2 terms.
Factoring x^2 - 5 using difference of perfect squares, resulting in complex and real roots.
General rule: for a polynomial of degree n, there should be n solutions, including real and complex numbers.
Summary of techniques: factoring by grouping, factoring by substitution, the Rational Zero Theorem, and synthetic division to solve polynomial equations.
Transcripts
in this tutorial i'm going to focus on
solving polynomial equations
so let's start with this one
x cubed plus
3x squared
minus 4x minus 12
is equal to zero
go ahead and solve this particular
equation
now
notice that
the ratio
of the first two coefficients is the
same as the ratio of the last two
so for example
one and three
has the same ratio
as negative four and negative twelve
negative twelve divided by negative four
is equal to three
and 3 divided by 1 is 3.
whenever you see that
you can factor by grouping
and in fact that's the best way or the
fastest way to get the answer
so let's take out the gcf in the first
two terms
the greatest common factor is x squared
x cubed divided by x squared is x
and three x squared divided by x squared
is three
now in the last two terms let's take out
the gcf as well
the greatest common factor is negative
four
negative four x divided by negative four
is x
negative twelve divided by negative four
is plus three
and so if you see this common factor
then
you're on or by track so let's take out
x plus three
so this whole thing divided by x plus
three
is x squared and this whole thing
divided by x plus three is negative four
now we're not done yet we can still
factor x squared minus four
it's a difference of perfect squares
the square root of x squared is x the
square root of four is two
one is going to be positive and the
other will be negative
so now
the polynomial expression
has been completely factored at this
point we can set each factor equal to
zero
and so we're going to get three answers
so x is equal to negative three
it's equal to negative two
and positive two as well
so that's how you can solve this
particular polynomial equation
now let's move on to our next example
so let's say that
x to the fourth
minus 5x squared
minus 36 is equal to zero
how can we solve this particular
polynomial equation
notice that
this equation is reducible
to a quadratic form so we can
factor it using substitution
let's say that a is equal to x squared
so that means that a squared is equal to
x to the fourth so let's replace x to
the fourth with a squared
and x squared with a so now we have a
trinomial with a leading coefficient of
one
so what two numbers multiply to negative
36 but add to negative five
so this is going to be
nine and four but negative 9 and
positive 4.
negative 9 times 4 is negative 36 but
negative 9 plus 4 is negative 5.
so to factor it's going to be a minus 9
and a plus 4.
so at this point i'm gonna replace a
with x squared so it's the same as x
squared minus nine
times x squared plus four
now you can factor x squared minus nine
it's going to be x plus three
and x minus 3 if you have a difference
of perfect squares
if you have a sum of perfect squares you
can factor using imaginary numbers
so it's going to be x plus 2i
and x minus 2i
so because it's
a polynomial of degree 4 there's gonna
be four answers but in this case
two of them are real numbers and the
other two are imaginary numbers
so x is equal to negative 3
3
negative 2i
and 2i so you can write it this way x is
equal to plus or minus 3 or
plus or minus 2i
so those are the four answers for this
problem
let's try this one x cubed minus three x
squared
minus six x plus eight
is equal to zero
find all the zeros of the function go
ahead and solve for the value of x
now we can't factor by grouping
because
the first two terms do not have the same
ratio as the last two
negative three divided by one
is not the same as eight divided by
negative six
which reduces to negative four thirds
so we can't factor by grouping
and we can't reduce it to quadratic form
so the last method that we can use
is we need to list all the possible
rational zeros using the rational zero
theorem and then use synthetic division
to find out the other zeros
so p
represents the factors of the constant
term eight
so factors of eight are one
two
four and eight
q
represents the factors of one the lean
coefficient
so
let's evaluate the function at x equals
one let's see if that is a possible
answer
one to the third minus three times one
squared minus six times one
plus eight let's see if it's equal to
zero
so one to the third is one three times
one squared is three
and then we have six plus eight
so one minus three is negative two
and negative two minus six is negative
eight negative eight plus eight is equal
to zero
because it's equal to zero
that means that this is one of the
answers x
is equal to one
now let's find the other answers
so let's use synthetic division
the coefficients are one
negative three negative six and eight
so let's bring down the one
one times one
is one
negative three plus one is negative two
and then one times negative two is
negative two
negative 6 plus negative 2 is negative 8
and then 1 times negative 8
is
negative 8 and 8 plus negative 8 is 0.
if this is not a 0 then
a mistake occurred
so this is going to be 1x squared which
is 1 degree lower than this one
minus 2x minus 8.
and let's set that equal to 0.
now let's see if we can factor this
expression two numbers that multiply to
8 but add to the middle coefficient of
negative two
are going to be negative four and
positive two
so to factor it's going to be x minus
four
times x plus two
so x is equal to four
and negative two
so we have three real solutions
four
negative two and one
here's another problem that we could try
x to the fourth power
minus seven x cubed
plus seven x squared
plus thirty five x
minus sixty
go ahead and solve this particular
polynomial equation
so using the rational zero theorem
the factors of 60 are going to be 1
2
3
4
5
6
seven and eight doesn't go into 60 but
10 goes into it
uh 12
15
30 and 60 and 20 as well
and then the leading coefficient is one
so there's a lot of factors
but let's start with one
let's see if f of one is equal to zero
so that's gonna be one to the fourth
power
minus seven times one cubed
plus seven times 1 squared
plus 35 times 1 minus 60.
so this is going to be 1 minus 7 plus 7
plus 35 minus 60.
negative 7 plus 7 is 0. and 35 minus 60
is negative 25
so negative 24 is not equal to zero so
this
doesn't work
let's try the next number two
so it's going to be 2 to the fourth
minus 7 times 2 to the third
plus 7
times 2 squared plus
35 times 2 minus sixty
so two to the fourth that's two times
two times two times two
two times two is four times two is eight
times two is sixteen
and then two to the third is eight
two squared is
four
thirty five times two is seventy
and then
seven times eight is fifty six
seven times four is twenty eight
and seventy minus sixty is ten
now sixteen minus fifty six
that's negative forty
and twenty eight plus 10
is 38
negative 40 plus 38 is negative two
which is not zero
so that doesn't work either
now let's try three
so three to the fourth power
minus seven times three to the third
plus seven
times three
squared plus thirty five times three
minus sixty
let's see if this gives us a zero three
to the fourth power is eighty one
three to the third is twenty seven
and three squared
is nine
now thirty five times three thirty times
three is ninety three times five is
fifteen ninety and fifteen is one of
five
now seven times twenty seven
seven times twenty is one forty
times 7 is 49 so 140 and 49 that's going
to be 189
7 times 9 is 63 and 105 minus 60 is 45.
so 81 minus 189 that's going to be
negative
108
and 63 plus 45 is positive 108. so this
adds up to zero
which means
x is equal to three
so that's one of the answers
now let's use synthetic division to find
the other ones
so the coefficients are 1
negative 7
7
35 and negative 60.
so let's bring down the one three times
one is three
negative seven plus three is negative
four
three times negative four is negative
twelve
seven plus negative twelve is negative
five
three times negative 5 is
negative 15
35 minus 15 is 20 and 3 times 20 is 60.
so if you get a 0 that's a good sign
so we started with degree 4 now it's
going to be a degree 3 function so it's
x cubed minus four x squared
minus five x plus twenty
so notice that we can factor by grouping
negative four divided by one is negative
four
twenty divided by negative five is also
negative 4.
so let's take out the gcf which is going
to be x squared
so x cubed divided by x squared is x
negative 4x squared divided by x squared
is negative 4.
and then let's take out a negative 5.
negative 5x divided by negative 5 is x
20 divided by negative 5 is
negative 4.
so it's going to be x minus 4
times x squared minus 5
and all of that
is equal to 0.
now let's factor x squared minus 5
using the difference of perfect squares
technique
so the square root of x squared is x the
square root of 5 is
the square root of 5.
and so there's going to be 4 answers in
total
so we could set x minus 4 equal to 0
x plus the square root of 5 equal to 0
and x minus the square root of 5 equal
to 0.
so x is equal to 4
negative square root 5
and positive square root five
so here's the first answer
the second one
the third one and the fourth
so you should have four answers for
a fourth degree polynomial and that's it
so now you know how to solve polynomial
equations by factoring by grouping
factoring by substitution
and using the rational zero theorem
along with synthetic division thanks
again for watching
you
Посмотреть больше похожих видео
Factoring Cubic Polynomials- Algebra 2 & Precalculus
Lec 37 - Zeroes of Polynomial Functions
Polinomial (Bagian 5) - Cara Menentukan Akar-akar Persamaan Polinomial
How To Factor Polynomials The Easy Way!
Synthetic Division of Polynomials
Solving Quadratic Equations by Extracting the Square Roots by @MathTeacherGon
5.0 / 5 (0 votes)