And I need an explanation because I solved it an my answer is (x +2x) ( 3x +2) and then the answer their is (x+1) (3x+2)

I don't understand this question 3x+5x+2 question

<h2>Factorising Trinomials Using Trial and Error</h2>
<p>What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.</p>
<p>Let’s factor the trinomial \(3{x}^{2}+5x+2\).</p>
<p>From our earlier work we expect this will factor into two binomials.</p>
<div>\(\begin{array}{c}\hfill 3{x}^{2}+5x+2\hfill \\ \hfill \left(\phantom{\rule{1em}{0ex}}\right)\left(\phantom{\rule{1em}{0ex}}\right)\hfill \end{array}\)</div>
<p>We know the first terms of the binomial factors will multiply to give us \(3{x}^{2}\). The only factors of \(3{x}^{2}\) are \(1x,3x\). We can place them in the binomials.</p>
<p><img alt="This figure has the polynomial 3 x^ 2 +5 x +2. Underneath there are two terms, 1 x, and 3 x. Below these are the two factors x and (3 x) being shown multiplied." src="https://data.ulearngo.com/assets/44793003-a462-4bda-b80e-2bc10a65e4f2"/></p>
<p>Check. Does \(1x·3x=3{x}^{2}\)?</p>
<p>We know the last terms of the binomials will multiply to 2. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of 2 are 1 and 2. But we now have two cases to consider as it will make a difference if we write 1, 2, or 2, 1.</p>
<p><img alt="This figure demonstrates the possible factors of the polynomial 3x^2 +5x +2. The polynomial is written twice. Underneath both, there are the terms 1x, 3x under the 3x^2. Also, there are the factors 1,2 under the 2 term. At the bottom of the figure there are two possible factorizations of the polynomial. The first is (x + 1)(3x + 2) and the next is (x + 2)(3x + 1)." src="https://data.ulearngo.com/assets/47f16ac6-b7d4-41a9-b338-4f731c629ba1"/></p>
<p>Which factors are correct? To decide that, we multiply the inner and outer terms.</p>
<p><img alt="This figure demonstrates the possible factors of the polynomial 3 x^ 2 + 5 x +2. The polynomial is written twice. Underneath both, there are the terms 1 x, 3 x under the 3 x ^ 2. Also, there are the factors 1, 2 under the 2 term. At the bottom of the figure there are two possible factorizations of the polynomial. The first is (x + 1)(3 x + 2). Underneath this factorization are the products 3 x from multiplying the middle terms 1 and 3 x. Also there is the product of 2 x from multiplying the outer terms x and 2. These products of 3 x and 2 x add to 5 x. Underneath the second factorization are the products 6 x from multiplying the middle terms 2 and 3 x. Also there is the product of 1 x from multiplying the outer terms x and 1. These two products of 6 x and 1 x add to 7 x." src="https://data.ulearngo.com/assets/0508273d-33bc-4c08-8dbb-ce8706177fcf"/></p>
<p>Since the middle term of the trinomial is 5<em>x</em>, the factors in the first case will work. Let’s FOIL to check.</p>
<div>\(\begin{array}{c}\left(x+1\right)\left(3x+2\right)\hfill \\ 3{x}^{2}+2x+3x+2\hfill \\ 3{x}^{2}+5x+2\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\)</div>
<p>Our result of the factoring is:</p>
<p>\(\begin{array}{}\\ 3{x}^{2}+5x+2\hfill \\ \left(x+1\right)\left(3x+2\right)\hfill \end{array}\)</p>
<div class="ns-school-emp">
<h3>Example: How to Factor Trinomials of the Form \(a{x}^{2}+bx+c\) Using Trial and Error</h3>
<div class="ns-school-defn">
<p>Factor completely: \(3{y}^{2}+22y+7\).</p>
<h3 data-type="title">Solution</h3>
<div data-type="title"><img alt="This table summarizes the steps for factoring 3 y ^ 2 + 22 y + 7. The first row states write the trinomial in descending order. The polynomial is written 3 y ^ 2 +22 y + 7." src="https://data.ulearngo.com/assets/335c8ec4-f897-4a10-ace2-197ed6b89db2"/><img alt="The second row states find all the factor pairs of the first term. The only pairs listed are 1 y, 3 y. Then, since there is only one pair, they are in the parentheses written (1 y ) and (3 y )." src="https://data.ulearngo.com/assets/6e380076-377e-426c-937a-50aea0a09352"/><img alt="The third row states “find all the factored pairs of the third term”. It also states the only factors of 7 are 1 and 7." src="https://data.ulearngo.com/assets/7f9cfa60-8f85-42da-9bdb-d7a66508bda5"/><img alt="The fourth row states test all the possible combinations of the factors until the correct product is found. The possible factors are shown (y + 1)(3 y + 7) and (y + 7)(3y + 1). Under each factor is the products of the outer terms and the inner terms. For the first it is 7y and 3y. For the second it is 21 y and y. The combination (y + 7)(3 y + 1) is the correct factoring." src="https://data.ulearngo.com/assets/319d9c5d-8e0d-4f75-80a5-c72c882fff78"/><img alt="The last row states to check by multiplying. The product of (y + 7)(3 y + 1) is shown as 3 y ^ 2 + 22 y + 7." src="https://data.ulearngo.com/assets/c3a48c7b-e074-40fa-8299-94bdcd819674"/></div>
</div>
</div>
<div class="ns-school-defn2">
<header>
<h3 data-type="title">Factor trinomials of the form \(a{x}^{2}+bx+c\) using trial and error.</h3>
</header>
<ol type="1">
<li>Write the trinomial in descending order of degrees.</li>
<li>Find all the factor pairs of the first term.</li>
<li>Find all the factor pairs of the third term.</li>
<li>Test all the possible combinations of the factors until the correct product is found.</li>
<li>Check by multiplying.</li>
</ol>
</div>
<p>When the middle term is negative and the last term is positive, the signs in the binomials must both be negative.</p>
<div class="ns-school-emp">
<h2>Example</h2>
<div class="ns-school-defn">
<div data-type="example">
<div>
<div>
<p>Factor completely: \(6{b}^{2}-13b+5\).</p>
</div>
<div>
<h3>Solution</h3>
<table summary="This figure shows the steps of factoring 6 b ^ 2 minus 13 b + 5. The first row states the trinomial is in descending order. The second row states find the factors of the first term. It has 1 b, 6 b and 2 b, 3 b under the first term. The third row states find the factors of the last term. consider the signs. Under the last term negative 1, negative 5 are listed.">
<tbody>
<tr>
<td style="width: 740px;">The trinomial is already in descending order.</td>
<td style="width: 197px;"><img alt="." src="https://data.ulearngo.com/assets/ea2cd214-7d84-4713-a9cc-44741a1721de"/></td>
</tr>
<tr>
<td style="width: 740px;">Find the factors of the first term.</td>
<td style="width: 197px;"><img alt="." src="https://data.ulearngo.com/assets/95e3c60e-007b-4834-afdc-705862256081"/></td>
</tr>
<tr>
<td style="width: 740px;">Find the factors of the last term. Consider the signs. Since the last term, 5 is positive its factors must both be positive or both be negative. The coefficient of the middle term is negative, so we use the negative factors.</td>
<td style="width: 197px;"><img alt="." src="https://data.ulearngo.com/assets/55c353aa-c11e-4eed-90c0-d394188b448e"/></td>
</tr>
</tbody>
</table>
<p>Consider all the combinations of factors.</p>
<table summary="The heading of this table is 6 b ^ 2 minus 13 b + 5. This table has two columns. The first column is labeled possible factors and the second column is labeled product. The first row has the factors (b – 1)(6 b – 5) and the product 6 b ^ 2 minus 11 b + 5. The second row has the factors (b – 5)(6 b minus 1) and has the product 6 b ^ 2 minus 31 b + 5. The third row has the factors (2b – 1)(3 b minus 5) and has the product 6 b ^ 2 minus13 b + 5. This row has an asterisk. The last row has the factors (2b – 5)(3b – 1) and has the product 6 b ^ 2 minus 17 b + 5.">
<thead>
<tr>
<th colspan="2" data-valign="top">\(6{b}^{2}-13b+5\)</th>
</tr>
<tr>
<th data-valign="top">Possible factors</th>
<th data-valign="top">Product</th>
</tr>
</thead>
<tbody>
<tr>
<td data-valign="top">\(\left(b-1\right)\left(6b-5\right)\)</td>
<td data-valign="top">\(6{b}^{2}-11b+5\)</td>
</tr>
<tr>
<td data-valign="top">\(\left(b-5\right)\left(6b-1\right)\)</td>
<td data-valign="top">\(6{b}^{2}-31b+5\)</td>
</tr>
<tr>
<td data-valign="top">\(\left(2b-1\right)\left(3b-5\right)\)</td>
<td data-valign="top">\(6{b}^{2}-13b+5\)*</td>
</tr>
<tr>
<td data-valign="top">\(\left(2b-5\right)\left(3b-1\right)\)</td>
<td data-valign="top">\(6{b}^{2}-17b+5\)</td>
</tr>
</tbody>
</table>
<p>\(\begin{array}{cccc}\text{The correct factors are those whose product}\hfill &amp; &amp; &amp; \\ \text{is the original trinomial.}\hfill &amp; &amp; &amp; \hfill \phantom{\rule{21em}{0ex}}\left(2b-1\right)\left(3b-5\right)\hfill \\ \text{Check by multiplying.}\hfill &amp; &amp; &amp; \\ \phantom{\rule{2.5em}{0ex}}\left(2b-1\right)\left(3b-5\right)\hfill &amp; &amp; &amp; \\ \phantom{\rule{2.5em}{0ex}}6{b}^{2}-10b-3b+5\hfill &amp; &amp; &amp; \\ \phantom{\rule{2.5em}{0ex}}6{b}^{2}-13b+5\phantom{\rule{0.2em}{0ex}}✓\hfill &amp; &amp; &amp; \end{array}\)</p>
</div>
</div>
</div>
</div>
</div>
<p>When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.</p>
<div class="ns-school-emp">
<h2>Example</h2>
<div class="ns-school-defn">
<div data-type="example">
<div>
<div>
<p>Factor completely: \(14{x}^{2}-47x-7\).</p>
</div>
<div>
<h3>Solution</h3>
<table summary="This figure has the steps to factor 14 x ^ 2 minus 47 x minus 7. The first row states the trinomial is in descending order and lists the trinomial. The second row states to find the factors of the first term. Under the first term the factors are written. They are 1 x, 14 x and 2 x, 7 x. The last row states to find the factors of the last term and consider the signs. under the last term are the factors 1, negative 7 and negative 1, 7.">
<tbody>
<tr>
<td style="width: 767px;">The trinomial is already in descending order.</td>
<td style="width: 171px;"><img alt="." src="https://data.ulearngo.com/assets/42d15893-9733-45f6-8cfa-07014fc1f315"/></td>
</tr>
<tr>
<td style="width: 767px;">Find the factors of the first term.</td>
<td style="width: 171px;"><img alt="." src="https://data.ulearngo.com/assets/a738e6df-d3dc-4bc9-b18f-15651e747652"/></td>
</tr>
<tr>
<td style="width: 767px;">Find the factors of the last term. Consider the signs. Since it is negative, one factor must be positive and one negative.</td>
<td style="width: 171px;"><img alt="." src="https://data.ulearngo.com/assets/4ada980c-656c-4d5a-b7ba-55205230de26"/></td>
</tr>
</tbody>
</table>
<p>Consider all the combinations of factors. We use each pair of the factors of \(14{x}^{2}\) with each pair of factors of \(-7.\)</p>
<table summary="This table has three columns. The first column is labeled factors of 14 x ^ 2. The second column is labeled “Pairs with” and the third column is labeled “factors of negative 7”. In the next row are the factors x, and 14 x. These factors pair with and 1, and negative 7, which are in the third column. In the next row are the factors x, 14 x which pair with negative 1 and 7 as indicated in the third column. In the next row down are the factors 2 x and 7 x, which pair with 1 and negative 7 as indicated in the third column. In the last row are the factors 2 x, 7 x, which pair with and negative 1 and 7 as indicated in the third column.">
<thead>
<tr>
<th data-valign="top">Factors of \(14{x}^{2}\)</th>
<th data-valign="top">Pair with</th>
<th data-valign="top">Factors of \(-7\)</th>
</tr>
</thead>
<tbody>
<tr>
<td data-valign="top">\(x\), \(14x\)</td>
<td data-valign="top"> </td>
<td data-valign="top">\(1\), \(-7\)<br/>
\(-7\), \(1\)<br/>
(reverse order)</td>
</tr>
<tr>
<td data-valign="top">\(x\), \(14x\)</td>
<td data-valign="top"> </td>
<td data-valign="top">\(-1\), \(7\)<br/>
\(7\), \(-1\)<br/>
(reverse order)</td>
</tr>
<tr>
<td data-valign="top">\(2x,7x\)</td>
<td data-valign="top"> </td>
<td data-valign="top">\(1\), \(-7\)<br/>
\(-7\), \(1\)<br/>
(reverse order)</td>
</tr>
<tr>
<td data-valign="top">\(2x,7x\)</td>
<td data-valign="top"> </td>
<td data-valign="top">\(-1\), \(7\)<br/>
\(7\), \(-1\)<br/>
(reverse order)</td>
</tr>
</tbody>
</table>
<p>These pairings lead to the following eight combinations.</p>
<img alt="This table has the heading 14 x ^ 2 – 47 x minus 7. This table has two columns. The first column is labeled “possible factors” and the second column is labeled “product”. The first column lists all the combinations of possible factors and the second column has the products. In the first row under “possible factors” it reads (x+1) and (14 x minus 7). Under product, in the next column, it says “not an option”. In the next row down, it shows (x minus 7) and (14 x plus 1). In the next row down, it shows (x minus 1) and (14 x plus 7). Next to this in the product column, it says “not an option.” The next row down under “possible factors”, it has the equation (x plus 7 and 14 x minus 1. Next to this in the product column it has 14 x ^2 plus 97 x minus 7. The next row down under possible factors, it has 2 x plus 1 and 7 x minus 7. Next to this under the product column, is says “not an option”. The next row down reads 2 x minus 7 and 7x plus 1. Next to this under the product column, it has 14 x ^2 minus 47 x minus 7 with the asterisk following the 7. The next row down reads 2 x minus 1 and 7 x plus 7. Next to this in the product column it reads “not an option”. The final row reads 2 x plus 7 and 7 x minus 1. Next to this in the product column it reads 14, x, ^ 2 plus 47 x minus 7. Next to the table is a box with four arrows point to each “not an option” row. The reason given in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor. That means that each of these combinations is not an option.”" src="https://data.ulearngo.com/assets/21b0a22c-11f5-47fe-9d0c-6b870bf7c8a7"/>
<p>\(\begin{array}{cccc}\text{The correct factors are those whose product is the}\hfill &amp; &amp; &amp; \hfill \phantom{\rule{19em}{0ex}}\left(2x-7\right)\left(7x+1\right)\hfill \\ \text{original trinomial.}\hfill &amp; &amp; &amp; \\ \text{Check by multiplying.}\hfill &amp; &amp; &amp; \\ \phantom{\rule{2.5em}{0ex}}\left(2x-7\right)\left(7x+1\right)\hfill &amp; &amp; &amp; \\ \phantom{\rule{2.5em}{0ex}}14{x}^{2}+2x-49x-7\hfill &amp; &amp; &amp; \\ \phantom{\rule{2.5em}{0ex}}14{x}^{2}-47x-7\phantom{\rule{0.2em}{0ex}}✓\hfill &amp; &amp; &amp; \end{array}\)</p>
</div>
</div>
</div>
</div>
</div>
<div class="ns-school-emp">
<h2>Example</h2>
<div class="ns-school-defn">
<div data-type="example">
<div>
<div>
<p>Factor completely: \(18{n}^{2}-37n+15\).</p>
</div>
<div>
<h3>Solution</h3>
<table summary="This figure lists the steps for factoring 18 n ^ 2 minus 37 n +15. The first row states the trinomial is in descending order and lists the trinomial. The second row states to find the factors of the first term. Under the first term the possible combinations of factors are listed as 1 n times18 n and 2 n times 9 n, and 3 n times 6 n. The third row states to find factors of the last term and consider the signs. Under the last term are the possible combinations of factors negative 1 times negative 15 and negative 3 times negative 5.">
<tbody>
<tr>
<td style="width: 703px;">The trinomial is already in descending order.</td>
<td style="width: 234px;">\(18{n}^{2}-37n+15\)</td>
</tr>
<tr>
<td style="width: 703px;">Find the factors of the first term.</td>
<td style="width: 234px;"><img alt="." src="https://data.ulearngo.com/assets/e3457e2f-7bef-4004-a41b-18024762ddc4"/></td>
</tr>
<tr>
<td style="width: 703px;">Find the factors of the last term. Consider the signs. Since 15 is positive and the coefficient of the middle term is negative, we use the negative facotrs.</td>
<td style="width: 234px;"><img alt="." src="https://data.ulearngo.com/assets/cfb9e6bb-d2a1-4b4e-bbba-d8c3698efb2c"/></td>
</tr>
</tbody>
</table>
<p>Consider all the combinations of factors.</p>
<img alt="This table has the heading 18 n ^ 2 – 37n + 15. This table has two columns. The first column is labeled possible factors and the second column is labeled product. The first column lists all the combinations of possible factors and the second column has the products. Eight rows list the product is not an option. There is a textbox giving the reason for no option. The reason in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor”. The row containing the factors (2n – 3)(9n – 5) with the product 18n^2 minus 37 n + 15 has an asterisk." src="https://data.ulearngo.com/assets/63f984a3-f79a-49fc-b18b-319c594fecbd"/>
<p>\(\begin{array}{cccc}\text{The correct factors are those whose product is}\hfill &amp; &amp; &amp; \hfill \phantom{\rule{19em}{0ex}}\left(2n-3\right)\left(9n-5\right)\hfill \\ \text{the original trinomial.}\hfill &amp; &amp; &amp; \\ \text{Check by multiplying.}\hfill &amp; &amp; &amp; \\ \phantom{\rule{2.5em}{0ex}}\left(2n-3\right)\left(9n-5\right)\hfill &amp; &amp; &amp; \\ \phantom{\rule{2.5em}{0ex}}18{n}^{2}-10n-27n+15\hfill &amp; &amp; &amp; \\ \phantom{\rule{2.5em}{0ex}}18{n}^{2}-37n+15\phantom{\rule{0.2em}{0ex}}✓\hfill &amp; &amp; &amp; \end{array}\)</p>
</div>
</div>
</div>
</div>
</div>
<p>Don’t forget to look for a GCF first.</p>
<div class="ns-school-emp">
<h2>Example</h2>
<div class="ns-school-defn">
<div data-type="example">
<div>
<div>
<p>Factor completely: \(10{y}^{4}+55{y}^{3}+60{y}^{2}\).</p>
</div>
<div>
<h3>Solution</h3>
<table summary="This figure points out the trinomial has a greatest common factor and to factor it first. The polynomial is 10 y ^ 4 + 55 y cubed + 60 y squared. The greatest common factor is 5 y squared. The next row states to factor the trinomial and has 5 y squared [2 y squared + 11 y +12] with possible factors y, 2 y under the first term and 1,12 and 2, 6 and 3, 4 under the last term.">
<tbody>
<tr>
<td> </td>
<td>\(10{y}^{4}+55{y}^{3}+60{y}^{2}\)</td>
</tr>
<tr>
<td>Notice the greatest common factor, and factor it first.</td>
<td>\(15{y}^{2}\left(2{y}^{2}+11y+12\right)\)</td>
</tr>
<tr>
<td>Factor the trinomial.</td>
<td><img alt="." src="https://data.ulearngo.com/assets/94d6aac2-3cbd-4bac-92d0-1fe9a64bbe78"/></td>
</tr>
</tbody>
</table>
<p>Consider all the combinations.</p>
<img alt="This table has the heading 2 y squared + 11 y + 12 This table has two columns. The first column is labeled “possible factors” and the second column is labeled “product”. The first column lists all the combinations of possible factors and the second column has the products. Four rows list the product is not an option. There is a textbox giving the reason for no option. The reason in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor”. The row containing the factors (y + 4)(2y + 3) with the product 2 y squared + 11 y + 12 has an asterisk." src="https://data.ulearngo.com/assets/dc0adf4b-3941-4e37-8e9f-b4577df64bd4"/>
<p>\(\begin{array}{cccc}\text{The correct factors are those whose product}\hfill &amp; &amp; &amp; \hfill \phantom{\rule{5em}{0ex}}5{y}^{2}\left(y+4\right)\left(2y+3\right)\hfill \\ \text{is the original trinomial. Remember to include}\hfill &amp; &amp; &amp; \\ \text{the factor}\phantom{\rule{0.2em}{0ex}}5{y}^{2}.\hfill &amp; &amp; &amp; \\ \text{Check by multiplying.}\hfill &amp; &amp; &amp; \\ \phantom{\rule{2.5em}{0ex}}5{y}^{2}\left(y+4\right)\left(2y+3\right)\hfill &amp; &amp; &amp; \\ \phantom{\rule{2.5em}{0ex}}5{y}^{2}\left(2{y}^{2}+8y+3y+12\right)\hfill &amp; &amp; &amp; \\ \phantom{\rule{2.5em}{0ex}}10{y}^{4}+55{y}^{3}+60{y}^{2}\phantom{\rule{0.2em}{0ex}}✓\hfill &amp; &amp; &amp; \end{array}\)</p>
</div>
</div>
</div>
</div>
</div>

77e14597-4d58-4399-be3f-b1bb5c28bc54

Factorising Trinomials Using Trial and Error

Mathematics studies measurement, relationships, and properties of quantities and sets, using numbers and symbols. Arithmetic, algebra, geometry, and calculus are popular topics in mathematics.

Mathematics

Learn how to factor polynomials, recognize different types of trinomials, and solve quadratic equations using techniques like factoring, the &#8220;ac&#8221; method, and the zero product property in this tutorial.


Factorising Trinomials Using Trial and Error

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