6.3: Factorization of trinomials of the form ax²+bx+c (2023)

Example \(\PageIndex{1}\)

Factor:

\(3x^{2}+7x+2\).

Solution:

Since the leading coefficient and the last term are prime, there is only one way to factor them both.

\(3=1\cdot 3\quad\text{e}\quad 2=1\cdot 2\)

First, write the factors of the first term, \(3x^{2}\), as follows:

\(3x^{2}+7x+2=(x\quad\color{Cerúleo}{?}\color{preto}{)(3x}\quad\color{Cerúleo}{?}\color{preto}{ )}\)

Both the middle and the last semester are positive; therefore, the factors of \(2\) are chosen as positive numbers. In this case, the only option is in which group to place these factors.

\((x+1)(3x+2)\quad\text{ou}\quad (x+2)(3x+1)\)

Determine which classification is correct by multiplying each expression.

\(\begin{alineado} (x+1)(3x+2)&=3x^{2}+2x+3x+2 \\ &=3x^{2}+5x+2\quad\color{rojo} {x}\\(x+2)(3x+1)&=3x^{2}+x+6x+2 \\ &=3x^{2}+7x+2\quad\color{Cerúleo}{\ marca de selección} \end{alinhado}\)

Please note that these products only differ in their transit times. Also note that the middle term is the sum of the inner and outer product as shown below:

respondedor:

\((x+2)(3x+1)\)

Example \(\PageIndex{2}\)

Factor:

\(12x^{2}+38x+20\).

Solution:

First consider the factors of the first and last terms.

\(\begin{matriz}{ccc}{12=1\cdot 12}&{\quad}&{20=1\cdot 20}\\{=2\cdot 6}&{\quad}&{=2 \cdot 10}\\{=3\cdot 4}&{\quad}&{=4\cdot 5} \end{matriz}\)

We are looking for products of factors whose sum is equal to the coefficient of the middle term \(38\). For brevity, the thought process is illustrated starting with the factors \(2\) and \(6\). The factorization begins at this point with the first term.

\(12x^{2}+38x+20=(2x\quad\color{Cerúleo}{?}\color{preto}{)(6x}\quad\color{Cerúleo}{?}\color{preto}{ )}\)

We are looking for factors of 20 that combine with the factors of 12 to give a mean of 38x

\(\begin{array}{lll} {Factors\:of\:20}&{Possible}&{factorization}\\{\color{Cerulean}{1\cdot 20}}&{(2x+1)( 6x+20)}&{\color{Cerulean}{meio\:term\Rightarrow 46x}}\\{\color{Cerulean}{1\cdot 20}}&{(2x+20)(6x+1)} &{\color{Cerulean}{meio\:term\Rightarrow 122x}}\\{\color{Cerulean}{2\cdot 10}}&{(2x+2)(6x+10)}&{\color{ Cerulean}{meio\:term\Rightarrow 32x}}\\{\color{Cerulean}{2\cdot 10}}&{(2x+10)(6x+2)}&{\color{Cerulean}{medium\ :term\Rightarrow 64x}}\\{\color{Cerulean}{4\cdot 5}}&{(2x+4)(6x+5)}&{\color{Cerulean}{middle\:term\Rightarrow 34x }}\\{\color{Cerulean}{4\cdot 5}}&{\color{OliveGreen}{(2x+5)(6x+4)}}&{\color{OliveGreen}{medium\:term\ Error 38x}\quad\color{Cerulean}{\checkmark}}\end{array}\)

Here the last combination gives an average of \(38x\).

respondedor:

\((2x+5)(6x+4)\)

Example \(\PageIndex{3}\)

Factor:

\(10x^{2}−23x+6\).

Solution

First consider the factors of the first and last terms.

\(\begin{matriz}{ccc}{10=1\cdot 10}&{\quad}&{6=1\cdot 6}\\{=2\cdot 5}&{\quad}&{=2 \cdot 3} \end{matriz}\)

We are looking for products of factors whose sum is equal to the coefficient of the middle term \(−23\). The factorization starts at this point with two sets of empty brackets:

\(10x^{2}-23x+6=(\quad)(\quad )\)

Since the last term is positive and the middle term is negative, we know that both factors of the last term must be negative. Here we list all the possible combinations with the factors \(10x^{2}=2x⋅5x\).

\(10x^{2}-23x+6=(2x\quad\color{Cerúleo}{?}\color{preto}{)(5x}\quad\color{Cerúleo}{?}\color{preto}{ )}\)

\(\begin{array}{ll}{(2x-1)(5x-6)}&{\color{Cerúleo}{meio\:term\Rightarrow -17x}}\\{(2x-6)(5x -1)}&{\color{Cerúleo}{meio\:termo\Seta Direita -32x}}\\{(2x-2)(5x-3)}&{\color{Cerúleo}{meio\:termo\ SetaDireita -16x}}\\{(2x-3)(5x-2)}&{\color{Cerulean}{middle\:term\Rightarrow -19x}} \end{array}\)

There is no combination that gives a mean of \(−23x\). We then turn to the factors of \(10x^{2}=10x⋅x\) and list all the possible combinations:

\(10x^{2}-23x+6=(10x\quad\color{Cerúleo}{?}\color{preto}{)(x}\quad\color{Cerúleo}{?}\color{preto}{ )}\)

\(\begin{array}{ll}{(10x-1)(x-6)}&{\color{Cerulean}{middle\:term\Rightarrow -61x}}\\{(10x-6)(x -1)}&{\color{Cerulean}{half\:right\Arrow Right -162x}}\\{(10x-2)(x-3)}&{\color{Cerulean}{half\:term\ SetaDireita -32x}}\\{\color{OliveGreen}{(10x-3)(x-2)}}&{\color{OliveGreen}{middle\:term\Rightarrow -23x}\quad\color{Cerulean} { \checkmark}} \end{matrix}\)

and we can write

respondedor:

\((10x-3)(x-2)\). The full review is left to the reader.

Example \(\PageIndex{4}\)

Factor:

\(5x^{2}+38x-16\).

Solution:

We start with the factors \(5\) and \(16\).

\(\begin{matriz}{cc}{}&{16=1\cdot 16}\\{5=1\cdot 5}&{=2\cdot 8}\\{}&{=4\cdot 4 } \end{matriz}\)

Since the leading coefficient is prime, we can start with the following:

\(5x^{2}+38x-16=(x\quad\color{Cerúleo}{?}\color{preto}{)(5x}\quad\color{Cerúleo}{?}\color{preto}{ )}\)

We are looking for products of factors 5 and 16 that could add up to 38.

\(\begin{array}{lll}{Factores\:de\:16}&{Mögliche}&{Produkte}\\{\color{Cerulean}{1\cdot 16}}&{1\cdot\color{ Cerulean}{1}\:\color{black}{e\:5}\cdot\color{Cerulean}{16}}&{\color{Cerulean}{products\Rightarrow\:1\:e\:80} }\\{\color{Azure}{1\cdot 16}}&{1\cdot \color{Azure}{16}\:\color{negro}{e\:5}\cdot\color{Azure}{ 1}}&{\color{Cerulean}{products\Rightarrow\:16\:e\:5}}\\{\color{Cerulean}{2\cdot 8}}&{1\cdot\color{Cerulean} {2}\:\color{schwarz}{e\:5}\cdot\color{Cerulean}{8}}&{\color{OliveGreen}{products\Rightarrow\:2\:e\:40}\quad \color{Cerulean}{\checkmark}}\\{\color{Cerulean}{2\cdot 8}}&{1\cdot\color{Cerulean}{8}\:\color{black}{e\:5 }\cdot\color{Cerulean}{2}}&{\color{Cerulean}{products\Rightarrow\:8\:e\:10}}\\{\color{Cerulean}{4\cdot 4}}& {1\cdot\color{Cerulean}{4}\:\color{schwarz}{e\:5}\cdot\color{Cerulean}{4}}&{\color{Cerulean}{products\Rightarrow\:4 \:e\:20}} \end{matriz}\)

Since the last term is negative, we must look for factors with opposite signs. Here we can see that the products 2 and 40 add up to 38 when they have opposite signs:

\(1\cdot (\color{Cerulean}{-2}\color{schwarz}{)+5\cdot}\color{Cerulean}{8}\color{cerulean}{=-2+40=38}\ )

So, use \(−2\) and \(8\) as factors of \(16\) and make sure that the inner and outer products are \(−2x\) and \(40x\):

respondedor:

\((x+8)(5x-2)\). The full review is left to the reader.

Exercise \(\PageIndex{1}\)

Factor:

\(12x^{2}-31x-30\)

respondedor

\((3x-10)(4x+3)\)

Example \(\PageIndex{5}\)

Factor:

\(9x^{2}+30xy+25y^{2}\).

Solution:

Find the factors of the first and last terms such that the sum of the inner and outer products is equal to the middle term.

\(\begin{matriz}{cc}{9x^{2}=1x\cdot 9x}&{25y^{2}=1y\cdot 25y}\\{=3x\cdot 3x}&{=5y\cdot 5y} \end{matriz}\)

Add the following products to get the middle term: \(3x⋅5y+3x⋅5y=30xy\).

\(\begin{alineado} 9x^{2}+30xy+25y^{2}&=(3x\quad )(3x\quad ) \\ &=(3x+5y)(3x+5y) \\ &= (3x+5a)^{2} \end{alinhado}\)

In this example we have a perfect quadratic trinomial. Review.

\(\begin{alineado} (3x+5y)^{2}&= 9x^{2}+2\cdot 3x\cdot 5y+25y^{2} \\ &=9x^{2}+30xy+25y ^{2}\quad\color{Cerúleo}{\checkmark} \end{alinhado}\)

respondedor:

\((3x+5a)^{2}\)

Exercise \(\PageIndex{2}\)

Factor:

\(16x^{2}−24xy+9y^{2}\).

respondedor

\((4x-3a)^{2}\)

Example \(\PageIndex{6}\)

Factor:

\(12x^{2}-27x+6\).

Solution:

Start by factoring the GCD.

\(12x^{2}-27x+6=3(4x^{2}-9x+2)\)

After factoring 3, the coefficients of the resulting trinomial are smaller and have fewer factors.

\(\begin{array}{cc}{4=\color{VerdeOliva}{1\cdot 4}}&{2=\color{VerdeOliva}{1\cdot 2}}\\{=2\cdot 2} &{}\end{matriz}\)

After some thought, we can see that the combination that gives the coefficient of the middle term is \(4(−2)+1(−1)=−8−1=−9\).

\(\begin{aligned}3(4x^{2}-9x+2)&=3(4x\quad\color{Cerulean}{?}\color{black}{)(x}\quad\color{Cerulean }{?}\color{schwarz}{)} \\ &=3(4x-1)(x-2) \end{alineado}\)

Review.

\(\begin{aligned} 3(4x-1)(x-2)&=3(4x^{2}-8x-x+2) \\ &=3(4x^{2}-9x+2) \\ &=12x^{2}-27x+6\quad\color{Sky}{\checkmark} \end{aligned}\)

The factor \(3\) is part of the factored form of the original expression; Be sure to include it in the answer.

respondedor:

\(3(4x-1)(x-2)\)

Example \(\PageIndex{7}\)

Factor:

\(−x^{2}+2x+15\).

Solution

In this example, the leading coefficient is \(−1\). Before you start factoring, factor \(−1\):

\(-x^{2}+2x+15=-1(x^{2}-2x-15)\)

At this point, factor the remaining trinomial as usual, remembering to write \(−1\) as a factor in your final answer. Since \(3 + (−5) = −2\), use \(3\) and \(5\) as factors of \(15\).

\(\begin{alinhado} -x^{2}+2x=15&=-1(x^{2}-2x-15) \\ &=-1(x\quad )(x\quad )\\ & =-(x+3)(x-5) \end{alinhado}\)

respondedor:

\(-1(x+3)(x-5)\). The examination is left to the reader.

Example \(\PageIndex{8}\)

Factor:

\(-60a^{2}-5a+30\)

Solution

The gcf of all terms is \(5\). In this case, however, factor \(−5\) because this produces a trinomial factor where the leading coefficient is positive.

\(-60a^{2}-5a+30=-5(12a^{2}+a-6)\)

Focus on the factors \(12\) and \(6\), which together give the mean coefficient \(1\).

\(\begin{array}{cc}{12=1\cdot 12}&{6=1\cdot 6}\\{=2\cdot 6}&{=\color{OliveGreen}{2\cdot 3} }\\{=\color{VerdeOliva}{3\cdot 4}}&{} \end{matriz}\)

After much thought we find that \(3⋅3−4⋅2=9−8=1\). Factor the remaining trinomial.

\(\begin{alineado} -60a^{2}-5a+30&=-5(12a^{2}+a-6) \\ &=-5(4a\quad )(3a\quad )\\& =-5(4a+3)(3a-2) \end{alinhado}\)

respondedor:

\(-5(4a+3)(3a-2)\). The examination is left to the reader.

Exercise \(\PageIndex{3}\)

Factor:

\(24+2x−x^{2}\).

respondedor

\(−1(x−6)(x+4)\)

Example \(\PageIndex{9}\)

Factor according to the AC method:

\(18x^{2}−21x+5\).

Solution:

Here \(a = 18, b = −21\) and \(c = 5\).

\(\begin{aligned}ac&=18(5) \\ &=90 \end{aligned}\)

Factor \(90\) and look for factors whose sum is \(−21\).

\(\begin{aligned} 90&=1(90) \\ &=2(45) \\ &=3(30) \\ &=5(18) \\ &=\color{OliveGreen}{6(15 )}\quad\color{Cerulean}{\checkmark} \\ &=9(10) \end{aligned}\)

In this case, the sum of the factors \(−6\) and \(−15\) is equal to the mean coefficient \(−21\). Then \(−21x=−6x−15x\), and we can write

\(18x^{2}\color{VerdeOliva}{-21x}\color{negro}{+5=18x^{2}}\color{VerdeOliva}{-6x-15x}\color{negro}{+5 }\)

Factor the equivalent expression by grouping.

\(\begin{alinhado} 18x^{2}-21x+5&=18x^{2}-6x-15x+5 \\ &=6x(3x-1)-5(3x-1) \\ &=( 3x-1)(6x-5) \end{alinhado}\)

respondedor:

\((3x-1)(6x-5)\)

Example \(\PageIndex{10}\)

Factor by the AC method: \(9x^{2}−61x−14\).

Solution:

Here \(a = 9, b = −61\) and \(c = −14\).

We factor \(-126\) as follows:

\(\begin{alineado} -126&=1(-126) \\ &=\color{OliveGreen}{2(-63)}\quad\color{Cerulean}{\checkmark} \\ &=3(-42 )\\&=6(-21)\\&=7(-18)\\&=9(-14) \end{alinhado}\)

The sum of the factors \(2\) and \(−63\) gives the average coefficient \(−61\). Replace \(−61x\) with \(2x−63x\):

\(\begin{aligned} 9x^{2}-61x-14&=9x^{2}+2x-63x-14\quad\color{Cerúleo}{Rearranja\:los\:términos.} \\ &=9x ^{2}-63x+2x-14\quad\color{Cerúleo}{Fator\:por\:agrupamento.}\\&=9x(x-7)+2(x-7) \\ &=(x -7)(9x+2) \end{alinhado}\)

respondedor:

\((x-7)(9x+2)\). The examination is left to the reader.

Homework \(\PageIndex{4}\) factor trinomials

Factor.

1. \(3x^{2}−14x−5\)
2. \(5x^{2}+7x+2 \)
3. \(2x^{2}+5x−3 \)
4. \(2x^{2}+13x−7 \)
5. \(2x^{2}+9x−5\)
6. \(7x^{2}+20x−3 \)
7. \(7x^{2}−46x−21 \)
8. \(3x^{2}+x−2 \)
9. \(5x^{2}+34x−7 \)
10. \(5x^{2}−28x−12 \)
11. \(9x^{2}−12x+4 \)
12. \(4x^{2}−20x+25 \)
13. \(49x^{2}+14x+1 \)
14. \(25x^{2}−10x+1 \)
15. \(2x^{2}+7x+16 \)
16. \(6x^{2}−19x−10 \)
17. \(27x^{2}+66x−16\)
18. \(12x^{2}−88x−15 \)
19. \(12a^{2}−8a+1 \)
20. \(16a^{2}−66a−27\)
21. \(9x^{2}−12xy+4y^{2} \)
22. \(25x^{2}+40x+16 \)
23. \(15x^{2}−26xy+8y^{2} \)
24. \(12a^{2}−4ab−5b^{2} \)
25. \(4x^{2}y^{2}+16xy−9 \)
26. \(20x^{2}y^{2}+4xy−7 \)
27. The area of ​​a rectangle is given by the function \(A(x)=3x^{2}−10x+3\), where \(x\) is measured in meters. Rewrite this function in factored form.
28. The area of ​​a rectangle is given by the function \(A(x)=10x^{2}−59x−6\), where \(x\) is measured in meters. Rewrite this function in factored form.

respondedor

1. \((x−5)(3x+1) \)

3. \((x+3)(2x−1) \)

5. \((x+5)(2x−1) \)

7. \((x−7)(7x+3) \)

9. \((x+7)(5x−1) \)

(Video) 6.3 Factoring Harder Trinomials

11. \((3x−2)^{2}\)

13. \((7x+1)^{2} \)

15. Director

17. \((3x+8)(9x−2)\)

19. \((6a-1)(2a-1)\)

21. \((3x−2y)^{2}\)

23. \((3x−4y)(5x−2y)\)

25. \((2xy−1)(2xy+9) \)

27. \(A(x)=(3x−1)(x−3)\)

Homework \(\PageIndex{5}\) Factorization of trinomials with common factors

Factor.

1. \(6x^{2}−20x−16 \)
2. \(45x^{2}+27x−18\)
3. \(20x^{2}−20x+5 \)
4. \(3x^{2}+39x−90\)
5. \(16x^{2}+26x−10 \)
6. \(54x^{2}−15x+6 \)
7. \(45x^{2}−45x−20 \)
8. \(90x^{2}+300x+250 \)
9. \(40x^{2}−36xy+8y^{2} \)
10. \(24a^{2}b^{2}+18ab−81 \)
11. \(6x^{2}y^{2}+46xy+28 \)
12. \(2x^{5}+44x^{4}+144x^{3}\)
13. \(5x^{3}−65x^{2}+60x\)
14. \(15a^{4}b^{2}−25a^{3}b−10a^{2}\)
15. \(6a^{4}b+2a^{3}b^{2}−4a^{2}b^{3}\)
16. \(20a^{3}b^{2}−60a^{2}b^{3}+45ab^{4}\)

respondedor

1. \(2(x−4)(3x+2) \)

3. \(5(2x−1)^{2}\)

5. \(2(8x^{2}+13x−5) \)

7. \(5(3x−4)(3x+1) \)

9. \(4(5x−2y)(2x−y)\)

11. \(2(xy+7)(3xy+2) \)

13. \(5x(x−12)(x−1) \)

15. \(2a^{2}b(3a−2b)(a+b)\)

Exercise \(\PageIndex{6}\) Factorization of trinomials with common factors

Factor \(−1\) and then factor it further.

1. \(−x^{2}−4x+21 \)
2. \(−x^{2}+x+12 \)
3. \(−x^{2}+15x−56 \)
4. \(−x^{2}+x+72 \)
5. \(−y^{2}+10y−25 \)
6. \(−y^{2}−16y−64 \)
7. \(36−9a−a^{2}\)
8. \(72−6a−a^{2}\)
9. \(32+4x−x^{2}\)
10. \(200+10x−x^{2}\)

respondedor

1. \(−1(x−3)(x+7) \)

3. \(−1(x−7)(x−8) \)

5. \(−1(y−5)^{2}\)

7. \(−1(a−3)(a+12) \)

9. \(−1(x−8)(x+4)\)

Exercise \(\PageIndex{7}\) Factorization of trinomials with common factors

Factor a negative common divisor first, then factor more if possible.

1. \(−8x^{2}+6x+9 \)
2. \(−4x^{2}+28x−49 \)
3. \(−18x^{2}−6x+4 \)
4. \(2+4x−30x^{2} \)
5. \(15+39x−18x^{2} \)
6. \(90+45x−10x^{2} \)
7. \(−2x^{2}+26x+28\)
8. \(−18x^{3}−51x^{2}+9x \)
9. \(−3x^{2}y^{2}+18xy^{2}−24y^{2} \)
10. \(−16a^{4}+16a^{3}b−4a^{2}b^{2} \)
11. The height in feet of a projectile fired from a turret is given by the function \(h(t)=−16t^{2}+64t+80\), where \(t\) is the number of seconds after represents the launch. Rewrite the given function in factored form.
12. The height in feet of a projectile fired from a turret is given by the function \(h(t)=−16t^{2}+64t+192\), where \(t\) is the number of seconds after represents the launch. Rewrite the given function in factored form.

respondedor

1. \(−(2x−3)(4x+3) \)

3. \(−2(3x−1)(3x+2) \)

5. \(−3(2x−5)(3x+1) \)

7. \(−2(x−14)(x+1) \)

9. \(−3y^{2}(x−4)(x−2) \)

11. \(h(t)=−16(t+1)(t−5) \)

Exercise \(\PageIndex{8}\) Factorization with the AC method

Factor using the AC method.

1. \(2x^{2}+5x−7 \)
2. \(3x^{2}+7x−10 \)
3. \(4x^{2}−25x+6 \)
4. \(16x^{2}−38x−5 \)
5. \(6x^{2}+23x−18 \)
6. \(8x^{2}+10x−25 \)
7. \(4x^{2}+28x+40 \)
8. \(−6x^{2}−3x+30 \)
9. \(12x^{2}−56xy+60y^{2}\)
10. \(20x^{2}+80xy+35y^{2}\)

respondedor

1. \((x−1)(2x+7) \)

3. \((x−6)(4x−1) \)

5. \((2x+9)(3x−2) \)

7. \(4(x+2)(x+5) \)

9. \(4(x−3y)(3x−5y)\)

Exercise \(\PageIndex{9}\) Discussion forum topics

1. Create your own trinomial of the form \(ax^{2}+bx+c\) that factors. Share it along with the solution in the discussion forum.
2. Write your own list of steps to factor a trinomial of the form \(ax^{2}+bx+c\) and share it in the discussion forum.
3. Create a trinomial of the form \(ax^{2}+bx+c\) that does not factor, and divide it along with the reason why it does not factor.

respondedor

1. Answers may vary

3. Answers may vary

(Video) Algebra - Factor a trinomial using the AC method