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enFactor trinomials mentally! Tips and tricks
https://mathalino.com/blog/factor-trinomials-mentally-tips-and-tricks
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<div class="collapse-text-text">Have you ever wondered how can we factor trinomials without writing anything on paper, i.e. <strong>mentally</strong>? The best way, unfortunately, is through trial-and-error method. However, to make factoring mentally faster, the trial-and-error must be an educated and systematic one.
<p>Here are some tips and tricks in factoring the trinomial $ax^{2}+bx+c$ mentally. Once you master the techniques in this blog, you can simplify expressions and solve equations that require factoring with "lightning" speed, and impress your friends.</p>
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<ul><li>If $a+b+c=0$, then one of the factors is $(x-1)$. The other one is $(ax-c)$.</li>
<li>If $a+c=b$, then one of the factors is $(x+1)$. The other one is $(ax+c)$.</li>
<li>If $b$ and $c$ are both positive, then each factor takes a $+$ sign.</li>
<li>If $b$ is negative and $c$ is positive, then each factor takes a $-$ sign.</li>
<li>If $c$ is negative, then one of the factors take a $+$ sign, and the other takes a $-$ sign.</li>
<ul><li>When selecting combinations of signs in this case, observe it from the resulting outer and inner terms, and the middle term of the original trinomial.</li>
</ul><li>When selecting for combinations of $ax^{2}$, try to eliminate some combinations based on the parity of the middle terms (including the resulting outer and inner terms). <strong>Parity</strong> means whether a number is even or odd.</li>
<li>When selecting for combinations of $c$, try to eliminate some combinations based on the following.</li>
<ul><li>If one of the factors is still factored by the GCF method, but the original trinomial is not factorable by the GCF method.</li>
<li>If either the resulting outer or inner term is too large compared to the middle term of the original polynomial.</li>
</ul><li>Do not do trial-and-error check on all cases, only check ones that you think will yield a middle term close to the middle term of the original trinomial. </li>
<li>Repeatedly practice finding combinations mentally. This will help you find factors and combinations faster, and later on, you will be able to find factors without exerting too much thinking at all.</li>
</ul></div>
<h3>Example 1</h3>
<p>Factor the trinomial $2x^{2}-x-6$.</p>
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<p></p><fieldset class="collapse-text-fieldset collapsible collapsed form-wrapper"><legend><span class="fieldset-legend">Here's the thought process when factoring this mentally.</span></legend><br /><div class="fieldset-wrapper">
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Because the first term is $2x^{2}$, we are certain that the factorization takes the form<br />
\[<br />
(2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)(x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)<br />
\]Taking note that the factors of $6$ are $1\times6$ and $2\times3$, we now do some deductions.
<ul><li>Since the third term $-6$ is negative, the two factors will take opposite signs of constants.</li>
<li>It is certainly not $(2x\square6)(x\square1)$. Otherwise, the first factor is still factorable by GCF as $2(x\square3)$. But we cannot take out $2$ via GCF from the original trinomial.</li>
<li>It is also certainly not $(2x\square1)(x\square6)$, because the outer terms multiply to $(2x)(6)=12x$, which is too large for the middle term $-x$ of the original trinomial. Also, notice that the inner terms multiply to $1(x)=x$, which is too small to reduce $12x$ to $-x$.</li>
</ul><p>This rules out $6=1\times6$. Hence, we are left with $2\times3$ as the factors of $6$.</p>
<ul><li>It is certainly not $(2x\square2)(x\square3)$ for the same reason as $(2x\square6)(x\square1)$.</li>
</ul><p>Hence, we are left with $(2x\square3)(x\square2)$. The outer terms multiply to $2x(2)=4x$, while the inner terms multiply to $3(x)=3x$.<br />
\[<br />
(\overbrace{2x\square\underbrace{3)(x}_{3x}\square2}^{4x})<br />
\] Because we need to get the middle term $-x$, and because the two factors take opposite signs, the only possibility is that $4x$ must be negative and $3x$ must be positive.<br />
\[<br />
-4x+3x=-x<br />
\] leading to<br />
\[<br />
(\overbrace{2x\square\underbrace{3)(x}_{3x}\square2}^{-4x})<br />
\] Therefore, the second square must be negative, and the first square must be positive.<br />
\[<br />
(\overbrace{2x+\underbrace{3)(x}_{3x}-2}^{-4x})<br />
\] Therefore, $2x^{2}-x-6=\boxed{(2x+3)(x-2)}$. Take note that all of these deductions are happening mentally.</p>
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<h3>Example 2</h3>
<p>Factor the trinomial $4x^{2}-19x+21$.</p>
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<p></p><fieldset class="collapse-text-fieldset collapsible collapsed form-wrapper"><legend><span class="fieldset-legend">Here's the thought process when factoring this mentally.</span></legend><br /><div class="fieldset-wrapper">
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Because the first term is $4x^{2}$, one possible combination is<br />
\[<br />
(2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)(2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)<br />
\] Because the leading coefficient of each factor is $2$, an even number, we are sure that the outer and inner terms are even, and consequently the resulting middle term is even. This is a contradiction, because the middle term of the original polynomial is $-19x$, which is an odd number. Hence, we reject this one and embrace the remaining possible combination<br />
\[<br />
(4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)(x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)<br />
\] Because the middle term is negative and the last term is positive, each of the factors must take negative signs.<br />
\[<br />
(4x-\square)(x-\square)<br />
\] The last term $21$ has the possible factors $1\times21$ and $7\times3$. We now do some deductions.
<ul><li>Definitely, $(x-1)$ is not one of the factors because the sum of the coefficients of the terms of the trinomial is $4-19+21=6\neq0$. So, $(4x-21)(x-1)$ is out.</li>
<li>Also, it is definitely not $(4x-1)(x-21)$, because the outer terms multiply to $4x(-21)=-84x$, which is too large.</li>
</ul><p>Hence, we are left with $21=7\times3$.</p>
<ul><li>It is not $(4x-3)(x-7)$, because the outer terms multiply to $(4x)(-7)=-28x$, which is too large compared to the middle term $-19x$.</li>
</ul><p>Therefore, we are left with the factorization $(4x-7)(x-3)$. A customary check of the middle term gives<br />
\[<br />
4x(-3)+(-7)(x)=-12x-7x=-19x<br />
\] which checks. Therefore, $4x^{2}-19x+21=\boxed{(4x-7)(x-3)}$. Take note that all of these deductions are happening mentally.
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<h3>Example 3</h3>
<p>Factor the trinomial $45x^{2}+79x-124$.</p>
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<p></p><fieldset class="collapse-text-fieldset collapsible collapsed form-wrapper"><legend><span class="fieldset-legend">Here's the thought process when factoring this mentally.</span></legend><br /><div class="fieldset-wrapper">
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This trinomial looks scary because of large coefficients. However, notice that the sum of the coefficients and constant terms is zero.<br />
\[<br />
45+79-124=0<br />
\] Therefore, one of the factors is automatically $(x-1)$. The other one is $({\color{blue}{45}}x{\color{red}{\,+\,124}})$. Therefore,<br />
\[<br />
{\color{blue}{45}}x^{2}+79x{\color{red}{\,-\,124}}=\boxed{(x-1)({\color{blue}{45}}x{\color{red}{\,+\,124}})}<br />
\] Observe carefully the colored numbers.
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<h3>Example 4</h3>
<p>Factor the trinomial $10x^{2}-17xy-27y^{2}$.
</p></div>
<p></p><fieldset class="collapse-text-fieldset collapsible collapsed form-wrapper"><legend><span class="fieldset-legend">Here's the thought process when factoring this mentally.</span></legend><br /><div class="fieldset-wrapper">
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Observe that the sum of the coefficients of the first and last terms equals the middle term.<br />
\[<br />
10+(-27)=-17<br />
\] Therefore, one of the factors is automatically $(x+1y)$. The other one is $({\color{blue}{10}}x{\color{red}{\,-\,27}}y)$. Therefore,<br />
\[<br />
{\color{blue}{10}}x^{2}-17xy{\color{red}{\,-\,27}}y^{2}=\boxed{(x+y)({\color{blue}{10}}x{\color{red}{\,-\,27}}y)}<br />
\]Again, observe carefully the colored numbers.
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<h3>Example 5</h3>
<p>Factor the trinomial $12x^{2}+7x-12$.
</p></div>
<p></p><fieldset class="collapse-text-fieldset collapsible collapsed form-wrapper"><legend><span class="fieldset-legend">Here's the thought process when factoring this mentally.</span></legend><br /><div class="fieldset-wrapper">
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The leading coefficient $12$ has factors $1\times12$, $2\times6$ and $3\times4$. The constant term $-12$ also has factors $1\times12$, $2\times6$ and $3\times4$, signs not included.
<ul><li>We will eliminate $12=2\times6$ immediately, because both $2$ and $6$ are even but the middle term $7x$ is odd.</li>
<li>We will eliminate $12=1\times12$ also, because the only possible combinations which will yield an odd middle term are $(x\square4)(12x\square3)$ and $(x\square12)(12x\square1)$.</li>
<ul><li>$(x\square4)(12x\square3)$ is out, because $(12x\square3)$ is still factorable by GCF, but the original trinomial is not.</li>
<li>$(x\square12)(12x\square1)$ is out, because the inner term gives $12(12x)=144x$, which is too large compared to the middle term $7x$.</li>
</ul></ul><p>So, we are left with $12=3\times4$.<br />
\[<br />
(3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)(4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)<br />
\]Recall that the constant term $-12$ also has factors $1\times12$, $2\times6$ and $3\times4$, signs not included.</p>
<ul><li>Because the constant term is negative, the two factors take opposite signs.</li>
<li>The factor with $4x$ can only take $\pm1$ and $\pm3$ as the constant term. Otherwise, this factor is still factorable via GCF, while the original trinomial is not, a contradiction.</li>
<ul><li>If the factor with $4x$ takes $\pm1$ as the constant term, then the factor with $3x$ takes $\pm12$ as the constant term, since $12=1\times12$. This is a contradiction, because the factor $(3x\square12)$ will now be factorable by GCF. Therefore, we eliminate $\pm1$.</li>
</ul></ul><p>Therefore, we are left with $\pm3$ as the constant term of the factor with $4x$, and $\pm4$ as the constant term of the factor with $3x$, since $12=3\times4$.<br />
\[<br />
(\overbrace{4x\square\underbrace{3)(3x}_{9x}\square4}^{16x})<br />
\]To get the middle term $7x$, we must have $16x+(-9x)=7x$.<br />
\[<br />
(\overbrace{4x\square\underbrace{3)(3x}_{-9x}\square4}^{16x})<br />
\]Therefore, the signs in the squares must be<br />
\[<br />
(\overbrace{4x-\underbrace{3)(3x}_{-9x}+4}^{16x})<br />
\]Hence, $12x^{2}+7x-12=\boxed{(4x-3)(3x+4)}$, and we are done.
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<h3>Practice problems</h3>
<p>Here are some more trinomials to practice what you have learned in this blog.</p>
<ol><li>$ 3x^2 - 13x - 10 $</li>
<li>$ 40x^2 - 79x + 39 $</li>
<li>$ 8x^2 + 26x + 15 $</li>
<li>$ 16x^2 - 33x - 49 $</li>
<li>$ 18x^2 + 9xy - 20y^2 $</li>
</ol></div>
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</div></div></div></div></div>Wed, 15 Sep 2021 09:42:25 +0000Engr Jaydee18754 at https://mathalino.comhttps://mathalino.com/blog/factor-trinomials-mentally-tips-and-tricks#comments