Factoring polynomials with a 2nd degree coefficient greater than $1$

Question

I'm learning how to factor polynomials, but I'm having a hard time understanding the approach when the 2nd degree coefficient is greater than $1$.

For example, when I begin to factor $12k^4 + 22k^3 - 70k^2$, I first break it down to $2k^2(6k^2 + 11k - 35)$.

I would think that I'd want to find two numbers that sum up to $11$ and have a product of $-35$, but instead I'm told we need to multiply $-35$ by $6$ so that I now have to find two numbers that sum up to $11$ and have a product of $-210$.

Can anyone help me understand why $-35$ is multiplied by the coefficient $6$? Why isn't $11k$ also multiplied by $6$?

Answer 1

That AC-method reduces to factoring a polynomial that is $\,\rm\color{#c00}{monic}\,$ (lead coeff $\color{#c00}{=1})$ as follows

$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) &\,=\,&\!\!\rm\: \color{#c00}{X^2} + b\:X + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\overbrace{ac,}^{\rm\qquad\ \ \ \ \ {\bf\large\ \ AC-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \ X = a\:x \\ \end{eqnarray}$$

In your case

$$ {\begin{eqnarray} f \, &\,=\,& \ \ \, 6 x^2+\ 11\ x\,\ -\ \ 35\\ \Rightarrow\,\ 6f\, &\,=\,&\!\,\ (6x)^2\! +11(6x)-210\\ &\,=\,& \ \ \ \color{#c00}{X^2}+\, 11\ X\,\ -\ 210,\,\ \ X\, =\, 6x\\ &\,=\,& \ \ (X+21)\ (X-\,10)\\ &\,=\,& \ (6x+21)\,(6x-10)\\ \Rightarrow\ \ f\:=\: \color{#0a0}{6^{-1}}\,(6f)\, &\,=\,& \ (2x+\,\ 7)\ (3x\,-5)\\ \end{eqnarray}}$$

In the final step we cancel $\,\color{#0a0}6\,$ by cancelling $\,3\,$ from the first factor, and $\,2\,$ from the second.

If we denote our factoring algorithm by $\,\cal F,\,$ then the above transformation is simply

$$\cal F f\, = a^{-1}\cal F\, a\,f\quad\,$$

Thus we've transformed by $ $ conjugation $\,\ \cal F = a^{-1} \cal F\, a\ \,$ the problem of factoring non-monic polynomials into the simpler problem of factoring monic polynomials. The same idea also works for higher degree polynomials, see this answer, which also gives links to closely-related ring-theoretic topics.

Answer 2

Your method

Suppose we had something simpler, with a quadratic coefficient ("leading coefficient") of $1$, like $k^2+2k-35$. Then we might expect to be able to factor it with factors whose leading coefficients are also $1$. Then the factorization would look like $\left(k+\square_1\right)\left(k+\square_2\right)$. Then when you multiply things out, you'll have $k^2+(\square_1+\square_2)k+\square_1\square_2$, so that we need the sum of the two numbers to be $2$ and the product to be $-35$. As you noted, this doesn't work the same way when the leading coefficient is not $1$.

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Adapting the method to this problem

You want to factor $6k^2+11k-35$.

Well, suppose you factor it with nice integer factors as $\left(\triangle_1k+\square_1\right)\left(\triangle_2k+\square_2\right)$. Then when you multiply things out, you'll have $\left(\triangle_1*\triangle_2\right)k^2+(\square_1*\triangle_2+\triangle_1*\square_2)k+\square_1*\square_2$. This means the triangle numbers have a product of $6$, the square numbers have a product of $-35$, and $\square_1*\triangle_2+\triangle_1*\square_2=11$.

We can assume the two triangle numbers are both positive (just take the negative of both factors otherwise) and that $\triangle_1>\triangle_2$ (just swap the two factors otherwise).

This leaves us with the following table of possibilities: $$\begin{matrix}\triangle_{1} & \triangle_{2} & \square_{1} & \square_{2} & \square_{1}*\triangle_{2}+\triangle_{1}*\square_{2} & = & 11?\\ 6 & 1 & 35 & -1 & 35*1+6*(-1) & 29 & \text{no}\\ 6 & 1 & -35 & 1 & & -29 & \text{no}\\ 6 & 1 & 7 & -5 & 7*1+6*(-5) & -23 & \text{no}\\ 6 & 1 & -7 & 5 & & 23 & \text{no}\\ 6 & 1 & 5 & -7 & 5*1+6*(-7) & -37 & \text{no}\\ 6 & 1 & -5 & 7 & & 37 & \text{no}\\ 6 & 1 & 1 & -35 & 1*1+6*(-35) & -\text{huge} & \text{no}\\ 6 & 1 & -1 & 35 & & \text{huge} & \text{no}\\ 3 & 2 & 35 & -1 & 35*2+3*(-1) & 67 & \text{no}\\ 3 & 2 & -35 & 1 & & -67 & \text{no}\\ 3 & 2 & 7 & -5 & 7*2+3*(-5) & -1 & \text{no}\\ 3 & 2 & -7 & 5 & & 1 & \text{no}\\ 3 & 2 & 5 & -7 & 5*2+3*(-7) & -11 & \text{no}\\ 3 & 2 & -5 & 7 & & 11 & \text{YES!}\\ 3 & 2 & 1 & -35 & 1*2+3*(-35) & -\text{big} & \text{no}\\ 3 & 2 & -1 & 35 & & \text{big} & \text{no} \end{matrix}$$

So the factorization is $6k^2+11k-35=\boxed{(3k-5)(2k+7)}$.

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Another method

Now, even ignoring the negatives, the table still had $8$ real cases in it, which is a bit of a pain. You can also factor this with the factor theorem. Basically, if there's a factor like $\left(k-\dfrac{5}{3}\right)$, then since $k=\dfrac{5}{3}$ would make that factor zero, it better make the whole polynomial zero as well. So if you know another way to find zeros/roots of the polynomial, you can use that to solve the problem.

As Battani implied, the quadratic formula is one method we can use. The zeros of the quadratic are therefore $$\dfrac{-11\pm\sqrt{11^{2}-4(6)(-35)}}{2(6)}=\dfrac{-11\pm\sqrt{121+840}}{2(6)}$$ $$=\dfrac{-11\pm\sqrt{900+61}}{12}=\dfrac{-11\pm\sqrt{30^{2}+(31-30)(31+30)}}{12}$$ $$ =\dfrac{-11\pm\sqrt{30^{2}+31^{2}-30^{2}}}{12}=\dfrac{-11\pm31}{12}=\dfrac{20}{12}\text{ and }\dfrac{-42}{12}=\dfrac{5}{3}\text{ and }\dfrac{-7}{2}$$

How does this help us factor things? Except for a constant, these must tell us the factors since we found the two zeros of the quadratic: $6k^2+11k-35=\square\left(k-\dfrac{5}{3}\right)\left(k-\left(\dfrac{-7}{2}\right)\right)$. We know it as to start with $6k^2$ after we multiply out, so we have $6k^2+11k-35=6\left(k-\dfrac{5}{3}\right)\left(k-\left(\dfrac{-7}{2}\right)\right)$. If we want nice integers instead of fractions, we can then take $3$ from the $6$ for the first factor and the $2$ from the $6$ for the second factor, to get $\boxed{\left(3k-5\right)\left(2k+7\right)}$.

Answer 3

We can factor your example as follows ,firslty,find roots of equation of $$6k^{ 2 }+11k-35=0\\ k=\frac { -11\pm 31 }{ 12 } =\frac { 5 }{ 3 } ,\frac { -7 }{ 2 } $$ so we can write it

$$2{ k }^{ 2 }\left( 6\left( k-\frac { 5 }{ 3 } \right) \left( k+\frac { 7 }{ 2 } \right) \right) =2{ k }^{ 2 }\left( 3k-5 \right) \left( 2k+7 \right) $$

Answer 4

To split the linear term of $6k^2 + 11k - 35$, we must find two numbers with product $6 \cdot -35 = -210$ and sum $11$. They are $21$ and $-10$. Hence, \begin{align*} 6x^2 + 11k - 35 & = \color{blue}{6}x^2 \color{green}{+ 21}x \color{green}{- 10}x \color{blue}{- 35} && \text{split the linear term}\\ & = 3x(2x + 7) - 5(2x + 7) && \text{factor by grouping}\\ & = (3x - 5)(2x + 7) && \text{extract the common factor} \end{align*} Notice that in the expression $\color{blue}{6}x^2 \color{green}{+ 21}x \color{green}{- 10}x \color{blue}{- 35}$, the product of the quadratic and constant coefficient is equal to the product of the two linear coefficients whose sum is the linear coefficient of $6x^2 + 11k - 35$, that is, $$(\color{blue}{6})(\color{blue}{-35}) = (\color{green}{21})(\color{green}{-10}) = -210$$ Let's examine the general case. Suppose $ax^2 + bx + c$ factors with respect to the rationals as $(rx + s)(tx + u)$. Then \begin{align*} ax^2 + bx + c & = (rx + s)(tx + u)\\ & = rx(tx + u) + s(tx + u)\\ & = \color{blue}{rt}x^2 + \color{green}{ru}x + \color{green}{st}x + \color{green}{su}\\ & = \color{blue}{rt}x^2 + (\color{green}{ru} + \color{green}{st})x + \color{blue}{su} \end{align*} Matching coefficients, we have $a = \color{blue}{rt}$, $b = \color{green}{ru + st}$, and $c = \color{blue}{su}$. Notice that the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients whose sum is $b$, that is, $$ac = (\color{blue}{rt})(\color{blue}{su}) = (\color{green}{ru})(\color{green}{st})$$ Thus, to split the linear term, we must find two numbers whose product is $ac$ and whose sum is $b$.

Answer 5

Factor $ak^2+bk-c$ (for fixed $a$, $b$, and $c$) first find numbers with product $-ac$, and sum $b$. Call these numbers $d$ and $f$. ($df$ = $-ac$ and $d+f$ = $b$)

$ax^2 + dx + fx - c$ is the linear form for variables.

Answer 6

(High School Level)

GUESS AND CHECK METHOD:

This method works well if you are just beginning and before too long you will start to see patterns and factor quickly. This speed comes in handy if you need to factor more than one expression in the same question or if you need to do a long list of factoring questions.

(1) GUESS: First, think of two possible factors of 6 and of 35. \begin{matrix} {6}{k}^{2} & +11k & -35 & \\ 2 & & 7 & \\ 3 & & 5 & \end{matrix}

(2) CHECK: Multiply the factors diagonally, then subtract the results to see if we get 11. \begin{matrix} {6}{k}^{2} & +11k & -35 & \\ 2 & & 7 & 3\times 7=21\\ 3 & & 5 & 2\times 5=10 \end{matrix} \begin{matrix}Check: 21-10=11\end{matrix}

        NOTE 1:  If the expression ends with a -, as with -35, we subtract in the check.  
    If it ends in a + then we add in the check.

        NOTE 2:  If the check did not work, we go back to step (1) and change the order of the factors
 or try different factors.

(3) SIGNS. If the expression ends with a -, we want the signs of the factors in the second column to be different, with one - and one +. We put the negative sign in the correct place so that the check still works. \begin{matrix} {6}{k}^{2} & +11k & -35 & \\ 2 & & 7 & 3\times 7=21\\ 3 & & -5 & 2\times -5=-10 \end{matrix} \begin{matrix}Check: 21-10=11\end{matrix}

If the expression ends in a +, we can skip this step because the signs will be the same.

(4) RESULT: Write down the result from left to right. \begin{matrix}{6}{k}^{2}+11k-35 = (2k+7)(3k-5)\end{matrix}

NOTE to teachers and tutors: This method is also excellent for making up questions for factoring practice. Simply make up the factoring table first (the factors that we guessed when we began the process) and use them to generate the desired polynomial.

Answer 7

SUMMARY: with $$ \color{blue}{b^2 - 4ac = \delta^2,} $$ take $$ \color{blue}{ a_1 = \gcd \left( a, \frac{(b + \delta)}{2} \right) ; \; \; \; \; \; a_2 = \frac{a}{a_1},} $$ then $$ \color{blue}{ a x^2 + b x y + c y^2 = \; \left(a_2x+ \left( \frac{b + \delta}{2a_1} \right) y \right) \; \; \left(a_1x+ \left( \frac{b - \delta}{2a_2} \right) y \right) \; } $$ in integers. If you do not want the variable $y,$ set $$ y=1. $$

EXAMPLE: $6 x^2 + 11 xy -35 y^2.$ $a = 6, b = 11, c = -35.$ $b^2 - 4ac = 961,$ $\delta = \sqrt {b^2 - 4ac} = 31.$ $(b+\delta)/ 2 = 21.$ $a_1 = \gcd(6,21) = 3.$ $(b+\delta)/ (2 a1) = 21/3 = 7.$ $a_2 = 6/3 = 2.$ $(b-\delta)/ 2 = -10.$ $(b-\delta)/ (2 a_2) = -10/2 = -5.$ $$ 6 x^2 + 11 xy -35 y^2 = (2x+7y)(3x-5y) $$ or $$ 6 x^2 + 11 x -35 = (2x+7)(3x-5) $$

EXAMPLE: $30 x^2 + 97 xy + 55 y^2.$ $a = 30, b = 97, c = 55.$ $b^2 - 4ac = 2809,$ $\delta = \sqrt {b^2 - 4ac} = 53.$ $(b+\delta)/ 2 = 75.$ $a_1 = \gcd(30,75) = \gcd(30, 15) = 15.$ $(b+\delta)/ (2 a1) = 75/15 = 5.$ $a_2 = 30/15 = 2.$ $(b-\delta)/ 2 = 22.$ $(b-\delta)/ (2 a_2) = 22/2 =11.$ $$ 30 x^2 + 97 xy + 55 y^2 = (2x+5y)(15x+11y) $$

EXAMPLE: $4 x^2 + 5 xy + y^2.$ $a = 4, b = 5, c = 1.$ $b^2 - 4ac = 9,$ $\delta = \sqrt {b^2 - 4ac} = 3.$ $(b+\delta)/ 2 = 4.$ $a_1 = \gcd(4,4) = 4.$ $(b+\delta)/ (2 a1) = 4/4 = 1.$ $a_2 = 4/4 = 1.$ $(b-\delta)/ 2 = 1.$ $(b-\delta)/ (2 a_2) = 1/1 =1.$ $$ 4 x^2 + 5 xy + y^2 = (x+y)(4x+y) $$

ORIGINAL: Either the one-variable function $$ a x^2 + b x + c $$ or the quadratic form $$ a x^2 + b x y + c y^2, $$ with integers $a,b,c,$ factor over the rational numbers if and only if the discriminant $$ \Delta = b^2 - 4 a c $$ is a square.

You are familiar with this because of the quadratic formula $$ \frac{-b \pm \sqrt \Delta}{2a} $$ which gives the roots $x$ for $ a x^2 + b x + c =0. $

Only a little changes when inserting the letter $y,$ giving $ a x^2 + b x y + c y^2. $ First, if both $a,c$ are $0,$ then we have $bxy$ which is already factored. So, let me show the traditional case, when $a \neq 0.$ Also let $$ \Delta = \delta^2, $$ say with integer $\delta \geq 0.$

$$ a x^2 + b x y + c y^2 = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + 4 a c y^2 \right) $$ $$ = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + b^2 y^2 - b^2 y^2 + 4 a c y^2 \right) $$ $$ = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + b^2 y^2 - (b^2 y^2 - 4 a c y^2) \right) $$ $$ = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + b^2 y^2 - (b^2 - 4 a c ) y^2 \right) $$ $$ = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + b^2 y^2 - \Delta y^2 \right) $$ $$ = \frac{1}{4a} \left( (2ax+by)^2 - \Delta y^2 \right) $$ $$ = \frac{1}{4a} \left( (2ax+by)^2 - \delta^2 y^2 \right) $$ $$ = \frac{1}{4a} \left( \; (2ax+by + \delta y) \; (2ax+by - \delta y) \; \right) $$ $$ = \frac{1}{4a} \left( \; (2ax+ (b + \delta) y) \; (2ax+ (b - \delta) y) \; \right) $$ Now, either $b,\delta$ are both even or both odd. Either way, we may absorb a factor of $4$ into $$ = \frac{1}{a} \left( \; (ax+ \frac{(b + \delta)}{2} y) \; (ax+ \frac{(b - \delta)}{2} y) \; \right) $$ Finally, since $$ \frac{(b + \delta)}{2} \frac{(b - \delta)}{2} = \frac{b^2 - \Delta}{4} = ac $$ is divisible by $a,$ by unique factorization we may write $$a = a_1 a_2$$ with $a_1$ dividing the first fraction and $a_2$ the second: in fact, we may simply take $$ a_1 = \gcd \left( a, \frac{(b + \delta)}{2} \right) ; \; \; \; \; \; a_2 = \frac{a}{a_1} $$ without paying attention to any prime factorizations of $a.$

We finally get $$ a x^2 + b x y + c y^2 = \; \left(a_2x+ \left( \frac{b + \delta}{2a_1} \right) y \right) \; \; \left(a_1x+ \left( \frac{b - \delta}{2a_2} \right) y \right) \; $$ in integers.