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learning goals

- Factor trinomials of the form \(ax^{2}+bx+c\).
- Factor trinomials with a common divisor.

## Factorization of trinomials of the form \(ax^{2}+bx+c\)

Factoring trinomials of the form \(ax^{2}+bx+c\) can be difficult because the middle term is affected by the factors \(a\) and \(c\). To illustrate, consider the following factored trinomial:

\(10x^{2}+17x+3=(2x+3)(5x+1)\)

We can multiply to verify that this is the correct factorization.

\(\begin{aligned} (2x+3)(5x+1)&=10x^{2}+2x+15x+3 \\ &=10x^{2}+17x+3\quad\color{Cerulean} {\check mark} \end{aligned}\)

As we saw previously, the product of the first terms of each binomial is equal to the first term of the trinomial. The middle term of the trinomial is the sum of the products of the outer and inner terms of the binomials. The product of the last term of each binomial is equal to the last term of the trinomial. Visually we have the following:

Generally,

\(\begin{aligned} \color{Cerulean}{a}\color{black}{x^{2}+}\color{Cerulean}{b}\color{black}{x+}\color{Cerulean}{ c}&= (px+m)(qx+n) \\ &=pqx^{2}+pnx+qmx+mn \\ &=\color{Cerúleo}{pq}\color{preto}{x^{ 2}+}\color{Cerúleo}{(pn+qm)}\color{preto}{x+}\color{Cerúleo}{mn} \end{alinhado}\)

this gives us

\[a=pq\quad\text{e}\quad b=pn+qm,\quad\text{onda}\quad c=mn\]

In summary, when the leading coefficient of a trinomial is other than \(1\), there is more to consider when determining the factors by trial and error. The key is to understand how the middle ground is reached. Multiply \((2x+5)(3x+7)\) and carefully follow the medium-term formation.

\(\begin{array}{ccc} {(\color{Cerúleo}{2x}\color{preto}{+}\color{OliveGreen}{5}\color{preto}{)(3x+7)=\ color{Cerulean}{2x}\color{black}{\cdot 3x}}}&{\underbrace{+\color{Cerulean}{2x}\color{black}{\cdot 7+}\color{OliveGreen}{ 5}\color{negro}{\cdot 3x}}}&{+\color{OliveGreen}{5}\color{negro}{\cdot 7}} \\ {}&{\color{Cerulean}{middle\ :término}}&{} \end{matriz}\)

\(\begin{alineado} &=6x^{2}+14x+15x+35 \\ &=6x^{2}+29x+35 \end{alineado}\)

If we think of the FOIL method of multiplying binomials, then the middle term is the sum of the inner product and the outer product. In this case \(14x+15x=29x\), as shown below:

So we need to look for products of the factors of the first and last terms whose sum is equal to the coefficient of the middle term. For example, to factor \(6x^{2}+29x+35\), look at the factors of \(6\) and \(35\).

\(\begin{array}{ccc}{6=1\cdot 6}&{\quad}&{35=1\cdot 35}\\{=\color{OliveGreen}{2\cdot 3}}&{ \quad}&{=\color{VerdeOliva}{5\cdot 7}} \end{matriz}\)

The combination that gives the coefficient of the middle term is \(2⋅7+3⋅5=14+15=29\). Make sure that the outer terms have coefficients \(2\) and \(7\) and the inner terms have coefficients \(5\) and \(3\). Use this information to factor the trinomial:

\(\begin{aligned} 6x^{2}+29x+35&=(2x\quad\color{Cerulean}{?}\color{black}{)(3x}\quad\color{Cerulean}{?}\ color{negro}{)} \\ &=(2x+5)(3x+7) \end{alineado}\)

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.

We can reduce a lot of the guesswork in factoring trinomials by considering all the factors of the first and last terms and their products.

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.

After much practice, the process described in the example above can be performed mentally.

Exercise \(\PageIndex{1}\)

Factor:

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

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

Given trinomials with multiple variables, the process is similar.

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

## Factor trinomials with common divisors

It is good practice to factor the GCD first, if it exists. This produces a trinomial factor with smaller coefficients. As we have seen, factoring trinomials with smaller coefficients requires much less effort. This often overlooked step is worth identifying.

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**:

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

It is advisable to consistently work with trinomials where the leading coefficient is positive.

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**-
\(−1(x−6)(x+4)\)

## Factorization according to the AC method

In this section we factor trinomials of the form \(ax^{2}+bx+c\) using the AC method described above.

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.

## Main Conclusions

- If a trinomial of the form \(ax^{2}+bx+c\) breaks down into the product of two binomials, then the coefficient of the middle term is the sum of certain products of factors of the first and last terms.
- If the trinomial has a greatest common factor, it's a good idea to first factor the LCD before trying to convert it to a product of binomials.
- If the leading coefficient of a trinomial is negative, it is good practice to factor that negative factor before attempting to factor the trinomial.
- Factoring trinomials of the form \(ax^{2}+bx+c\) requires a lot of practice and patience. Taking the time to become proficient through a lot of practice is extremely important.

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

Factor.

- \(3x^{2}−14x−5\)
- \(5x^{2}+7x+2 \)
- \(2x^{2}+5x−3 \)
- \(2x^{2}+13x−7 \)
- \(2x^{2}+9x−5\)
- \(7x^{2}+20x−3 \)
- \(7x^{2}−46x−21 \)
- \(3x^{2}+x−2 \)
- \(5x^{2}+34x−7 \)
- \(5x^{2}−28x−12 \)
- \(9x^{2}−12x+4 \)
- \(4x^{2}−20x+25 \)
- \(49x^{2}+14x+1 \)
- \(25x^{2}−10x+1 \)
- \(2x^{2}+7x+16 \)
- \(6x^{2}−19x−10 \)
- \(27x^{2}+66x−16\)
- \(12x^{2}−88x−15 \)
- \(12a^{2}−8a+1 \)
- \(16a^{2}−66a−27\)
- \(9x^{2}−12xy+4y^{2} \)
- \(25x^{2}+40x+16 \)
- \(15x^{2}−26xy+8y^{2} \)
- \(12a^{2}−4ab−5b^{2} \)
- \(4x^{2}y^{2}+16xy−9 \)
- \(20x^{2}y^{2}+4xy−7 \)
- 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.
- 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 Trinomials11. \((3x−2)^{2}\)

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.

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

**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.

- \(−x^{2}−4x+21 \)
- \(−x^{2}+x+12 \)
- \(−x^{2}+15x−56 \)
- \(−x^{2}+x+72 \)
- \(−y^{2}+10y−25 \)
- \(−y^{2}−16y−64 \)
- \(36−9a−a^{2}\)
- \(72−6a−a^{2}\)
- \(32+4x−x^{2}\)
- \(200+10x−x^{2}\)

**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.

- \(−8x^{2}+6x+9 \)
- \(−4x^{2}+28x−49 \)
- \(−18x^{2}−6x+4 \)
- \(2+4x−30x^{2} \)
- \(15+39x−18x^{2} \)
- \(90+45x−10x^{2} \)
- \(−2x^{2}+26x+28\)
- \(−18x^{3}−51x^{2}+9x \)
- \(−3x^{2}y^{2}+18xy^{2}−24y^{2} \)
- \(−16a^{4}+16a^{3}b−4a^{2}b^{2} \)
- 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.
- 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.

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

**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

- Create your own trinomial of the form \(ax^{2}+bx+c\) that factors. Share it along with the solution in the discussion forum.
- Write your own list of steps to factor a trinomial of the form \(ax^{2}+bx+c\) and share it in the discussion forum.
- 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

## FAQs

### How do I factoring a trinomial? ›

**Conclusion: How to Factor a Trinomial**

- Step 1: Identify the values for b and c.
- Step 2: Find two numbers that ADD to b and MULTIPLY to c.
- Step 3: Use the numbers you picked to write out the factors and check.

**What is the first step in factoring a trinomial of the form ax2 bx c? ›**

Step 1: **Look for a GCF and factor it out first**. Step 2: Multiply the coefficient of the leading term a by the constant term c. List the factors of this product (a • c) to find the pair of factors, f1 and f2, that sums to b, the coefficient of the middle term.

**How do you factor a trinomial without a GCF? ›**

**If you have four terms with no GCF, then try factoring by grouping.**

- Step 1: Group the first two terms together and then the last two terms together.
- Step 2: Factor out a GCF from each separate binomial.
- Step 3: Factor out the common binomial.

**Can all trinomials of the form x2 bx c be factored true or false? ›**

If a trinomial of the form x2+bx+c factors into the product of two binomials, then the coefficient of the middle term is the sum of factors of the last term. **Not all trinomials can be factored as the product of binomials with integer coefficients**.

**How do you solve factorization? ›**

To solve a quadratic equation by factoring, **arrange all the terms on one side of the equation so the other side equals 0, factor the expression, set each factor equal to 0 and solve each equation**.

**What are the rules of trinomials? ›**

The rules are as follows: **If all terms of the trinomial are positive, then all terms of the binomials will be positive**. If the last term of the trinomial is negative but the middle term and the first term are positive, then one term of the binomial will be negative and the other will be positive.

**What is a trinomial in math? ›**

A trinomial is **an algebraic expression that has three terms in it**. The examples of trinomials are: x + y + 7.

**What are the three steps for solving a quadratic equation by factoring? ›**

**Step 1: Obtain zero on one side and then factor.** **Step 2: Set each factor equal to zero.** **Step 3: Solve each of the resulting equations**. This technique requires the zero factor property to work so make sure the quadratic is set equal to zero before factoring in step 1.

**What are the steps in squaring a trinomial? ›**

**Origins of the Formula**

- Multiply the first term of the first factor by each of the terms in the second factor.
- Multiply the second term of the first factor by each of the terms in the second factor.
- Continue this pattern for each of the terms in the first factor, then add up all of the products.

**How do you solve quadratic equations by factoring for beginners? ›**

**To solve an quadratic equation using factoring :**

- 1 . Transform the equation using standard form in which one side is zero.
- 2 . Factor the non-zero side.
- 3 . Set each factor to zero (Remember: a product of factors is zero if and only if one or more of the factors is zero).
- 4 . Solve each resulting equation.

### What is an example of trinomial equation? ›

Examples of a trinomial expression: **x + y + z** is a trinomial in three variables x, y and z. 2a^{2} + 5a + 7 is a trinomial in one variables a. xy + x + 2y^{2} is a trinomial in two variables x and y.

**Why can't I factor a trinomial? ›**

Note: Some trinomials cannot be factored. **If none of the pairs total b, then the trinomial cannot be factored**. Example 1: Factor x^{2} + 5x + 6. Pairs of numbers which make 6 when multiplied: (1, 6) and (2, 3).

**How do you factor a trinomial using the box method? ›**

To factor a trinomial using the box method, first place the first term in the first box and the last term in the last box. Then multiply the first and last box. Find two terms which are the product of the first and last box, and which add up to be the middle term. Place these terms into the remaining boxes.

**What is the factor form of X² y²? ›**

The difference of squares, x² – y², can be factored as the product of the sum and difference of two terms, or **x² – y² = (x + y)(x – y)**. Example: 2x2 – 18 = 2(x² – 9), = 2(x – 3)(x + 3).