Factoring by Substitution PDF

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Algebra Worksheet Solutions of Factoring Polynomials (2) Solutions: 1. Factoring (x − 3)2 − 6(x − 3) + 8 Step 1. Let 𝑦 = x − 3, so rewrite the given polynomial in 𝑦 2 − 6𝑦 + 8 Step 2. Factoring 𝑦 2 − 6𝑦 + 8 = (𝑦 − 2)(𝑦 − 4) Step 3. Substitute 𝑦 = x − 3 back: (𝑥 − 3 − 2)(𝑥 − 3 − 4) Step 4. Simplify: (𝑥 − 5)(𝑥 − 7) The solution is (𝐱 − 𝟑)𝟐 − 𝟔(𝐱 − 𝟑) + 𝟖 = (𝒙 − 𝟓)(𝒙 − 𝟕)

2. Factoring 4(2 − 5x)2 + 8(2 − x) + 3 Let 𝑦 = 2 − 5x, so rewrite the given polynomial in 4𝑦 2 + 8𝑦 + 3 Step 2. Factoring 4𝑦 2 + 8𝑦 + 3 = 4𝑦 2 + 6𝑦 + 2𝑦 + 3 = (4𝑦 2 + 6𝑦) + (2𝑦 + 3) = 2𝑦(2𝑦 + 3) + (2𝑦 + 3) = (2𝑦 + 1)(2𝑦 + 3) Step 3. Substitute 𝑦 = 2 − 5x back: [2(2 − 5𝑥) + 1] [2(2 − 5𝑥) + 3] Step 4. Simplify: [2(2 − 5𝑥) + 1] [2(2 − 5𝑥) + 3] = (5 − 10𝑥)(7 − 10𝑥) The solution is: 𝟒(𝟐 − 𝟓𝐱)𝟐 + 𝟖(𝟐 − 𝐱) + 𝟑 = (𝟓 − 𝟏𝟎𝒙)(𝟕 − 𝟏𝟎𝒙)

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Algebra Worksheet Solutions of Factoring Polynomials (2) 3. Factoring 14(x + 1)2 − 35(x + 1) + 21 𝑃𝑢𝑙𝑙 𝑐𝑜𝑚𝑚𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 7: 7[2(x + 1)2 − 5(x + 1) + 3] Substitution: 𝑦 = x + 1: 7[2𝑦2 − 5𝑦 + 3] = 7[2𝑦 2 − 2𝑦 − 3𝑦 + 3] Grouping: 7[2𝑦2 − 2𝑦 − 3𝑦 + 3] = 7[(2𝑦2 − 2𝑦) − (3𝑦 − 3)] Pull common factors: 7[(2𝑦 2 − 2𝑦) − (3𝑦 − 3)] = 7[2𝑦(𝑦 − 1) − 3(𝑦 − 1)] Pull common factor (𝑦 − 1): 7[(𝑦 − 1)(2𝑦 − 3)] Substitute 𝑦 = x + 1 back: 7(𝑥 + 1 − 1)[2(𝑥 + 1) − 3] = 7𝑥(2𝑥 − 1) 𝐓𝐡𝐞 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐢𝐬: 𝟏𝟒(𝐱 + 𝟏)𝟐 − 𝟑𝟓(𝐱 + 𝟏) + 𝟐𝟏 = 𝟕𝒙(𝟐𝒙 − 𝟏) 4. Factoring 30(x + 1)2 + (x + 1) − 1 Substitution: 𝑦 = x + 1: 30 y2 + y − 1 = 30 y 2 + 6y − 5y − 1 Grouping: 30 y2 − 5y + 6y − 1 = (30 y 2 − 5y) + (6y − 1) Pull common factor: 5𝑦(6y − 1) + (6y − 1) = (6y − 1)(5y + 1) Substitute 𝑦 = x + 1 back: [6(x + 1) − 1][5(x + 1) + 1] Simplify and the solution is: (6x + 5)(5x + 6) 𝐓𝐡𝐞 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐢𝐬: 𝟑𝟎(𝐱 + 𝟏)𝟐 + (𝐱 + 𝟏) − 𝟏 = (𝟔𝐱 + 𝟓)(𝟓𝐱 + 𝟔)

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Algebra Worksheet Solutions of Factoring Polynomials (2) 5. Factoring 8(x + a)2n + 6(x + a)n – 9 Let (x + a) = 𝑦 So, 8𝑦 2𝑛 + 6𝑦𝑛 − 9 = 8𝑦 2𝑛 − 6𝑦 𝑛 + 12𝑦 𝑛 − 9 = (8𝑦 2𝑛 − 6𝑦 𝑛 ) + (12𝑦 𝑛 − 9) = 2𝑦 𝑛 (4𝑦𝑛 − 3) + 3(4𝑦𝑛 − 3)

= (4𝑦𝑛 − 3)(2𝑦𝑛 + 3) The solution = [𝟒(𝒙 + 𝒂 )𝒏 − 𝟑][𝟐(𝒙 + 𝒂)𝒏 + 𝟑]

6. Factoring 9(3a − 2)2 − 16(𝑏 + 1)2 Apply x2 − 𝑦 2 = (𝑥 + 𝑦)(𝑥 − 𝑦)𝑓𝑜𝑟𝑚𝑢𝑙𝑎𝑟: 9(3a − 2)2 − 16(𝑏 + 1)2 = [3(3𝑎 − 2)]2 − [4(𝑏 + 1)]2 = [3(3𝑎 − 2) + 4(𝑏 + 1)][3(3𝑎 − 2) − 4(𝑏 + 1)] The solution = (𝟗𝒂 + 𝟒𝒃 − 𝟐)(𝟗𝒂 − 𝟒𝒃 − 𝟏𝟎)

7. Factoring 4(x + 3)2 − 25(𝑦 + 2)2 4(x + 3)2 − 25(𝑦 + 2)2 = [2(𝑥 + 3) + 5(𝑦 + 2)][2(𝑥 + 3) − 5(𝑦 + 2)] The solution = (𝟐𝒙 + 𝟓𝒚 + 𝟏𝟔)(𝟐𝒙 − 𝟓𝒚 − 𝟒)

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Factoring Higher Powers MathBitsNotebook.com Topical Outline | Algebra 1 Outline | MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts

Sometimes factoring problems with higher degree powers can be rewritten as a simpler factoring problem. This first example replaces sections of the problem with a new variable. This concept is referred to as factoring using substitution.

Factor: a4 + a2 - 12 At first glance, this problem doesn't look like anything we have seen. But upon closer examination, this is really a very simple problem that we already know how to solve. The trick is to look at the problem in a slightly different manner.

a4 + a2 - 12 = (a2)2 + (a2) - 12 If you now replace a2 with x (or your favorite variable letter), the problem becomes very familiar.

Let a2 = x and substitute.

x2 + x - 12 = (x + 4)(x - 3)

Factor the new form: Re-substitute x = a2.

(a2)2 + (a2) - 12 = x2 + x - 12

(x + 4)(x - 3) = (a2 + 4)(a2 - 3) ANSWER

This second example creates a new problem by finding a common factor.

Factor: 2x5 - 9x4 - 5x3 Look carefully! Each of these terms has a factor of x3. Factoring out x3 may create a problem which can be factored further.

x3(2x2 - 9x - 5) Factor the remaining trinomial using any method you wish.

Factor:

x3(2x2 - 9x - 5) = x3(2x + 1)(x - 5) ANSWER

This third example has several solution methods.

Factor: m8 - 16

Since m8 and 16 are perfect squares, and this problem deals with subtraction, we are looking at factoring the difference of perfect squares. (m4 - 4)(m4 + 4) But the first of the two factors is ANOTHER difference of perfect squares. Repeat the process.

(m2 - 2)(m2 + 2)(m4 + 4) ANSWER This problem can also be solved using the substitution approach.

(m4)2 - 16 Let m4 = x. (m4)2 - 16 = (x)2 - 16 Factor: (x)2 - 16 = (x - 4)(x + 4) Re-substitute: (x - 4)(x + 4) = (m4 - 4)(m4 + 4) Repeat process: (m2)2 - 4 Let m2 = b. (m2)2 - 4 = (b)2 - 4 Factor: (b)2 - 4 = (b - 2)(b + 2) Re-substitute: (b - 2)(b + 2) = (m2 - 2)(m2 + 2) Total answer: (m2 - 2)(m2 + 2)(m4 + 4) ANSWER

This is a tough one!! What happens if there is a variable in the exponent?

Factor: x 2p + 2x p - 24 You have to be a bit more creative to see this solution. The secret is in the exponents. x p • x p = x 2p making x 2p the square of x p. A substitution will solve our problem.

(x p )2 + 2(x p ) - 24 Let x p = a and substitute. Factor the new form: Re-substitute a = x p.

(x p)2 + 2(x p) - 24 = a2 + 2a - 24 a2 + 2a - 24 = (a - 4)(a + 6)

(a - 4)(a + 6) = (x p - 4)(x p+ 6) ANSWER

NOTE: Without further information about "p", we do not know if further factoring is needed. If p were a positive even exponent, (x p - 4) could be the difference of two squares and could be factored further. However, without that information, we stop at our labeled answer.

NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". Topical Outline | Algebra 1 Outline | MathBitsNotebook.com | MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts Copyright 2012-2019 MathBitsNotebook.com All Rights Reserved

Factoring by Substitution Generally, factoring polynomials means separating a polynomial into its component polynomials. Sometimes, when polynomi particularly complicated, it is easiest to substitute a simple term and factor down. Sometimes this looks like the example belo where a multi-variable polynomial is substituted for a single variable,

in this case, which stands in for the single polynomial

overall polynomial is factored down. This is a quick technique to use when faced with complicated polynomials, particularly if subsets of that polynomial repeat. F instance, in the example below,

repeats twice in the equation. Sometimes manipulation of the overall polynomial is ne

to see this repetition. It is also useful when faced with complicated components like roots, imaginary numbers, and other non non-integer numbers. EXAMPLE

Factor

.

Let

. Then substituting

in for

for ease of computation, we get

EXAMPLE

Factor

Let

.

=

. Then substituting

in for

, we get

TRY IT YOURSELF

If

is a factor of

, then find the value of .

1 3 2 0

Challenging Example Questions

The following is an example of substitution of imaginaries: EXAMPLE

Solve for all roots of the polynomial

We begin by substituting

(be careful of +/- signs), which leaves us with

Looks much more factorable, doesn't it? Now it's just grouping, grouping, and more grouping. (If you want, you can use this

Using some logic, since there's a gap in powers (

followed by ), there should be a difference of squares. By remainder

theorem, since 5 is a solution, we can check -5.

Some synthetic division yields

Re-substituting

, we arrive at

Substitution can also be used to minimize and maximize functions. EXAMPLE

Find the minimum value of

.

The given expression can be rewritten as

. Substituting

. Completing the square gives

which implies that the minimum value of the given expression is

Substitution can also be used for irrational coefficients. EXAMPLE

Find the sum of all positive roots of

Substituting

gives

.

, this is equal to

Since we are interested in positive roots,

which implies that the sum of all positive roots of the given expression is

Cite as: Factoring by Substitution. Brilliant.org. Retrieved 22:49, February 28, 2019, from https://brilliant.org/wiki/factor-polynomials-by-substitution/...