Factoring Quadratics

Factoring quadratics is a method that allows us to simplify quadratic expressions and solve equations. Common cases include factoring trinomials and factoring differences of squares.

A quadratic expression may be written as a sum, \(x^2+7x+12,\) or as a product \((x+3)(x+4),\) much the way that 14 can be written as a product, \(7\times 2,\) or a sum, \(6+8.\) Factoring a trinomial is the process of rewriting a sum as a product.

Quadratic expressions may be written in standard form as \(ax^2+bx+c.\)

Let's begin by factoring trinomials with \(a=1,\) such as \(x^2+8x+15.\)

First, we need to find the product \(ac:\) \(ac = (1)(15) = 15.\)

Next, we need to find a factor pair of \(ac\) that sums to \(b.\) So we need a factor pair of 15 that sums to 8. The factor pair of 3 and 5 sums to 8.

Next, we need to rewrite the "\(b\)-term" of our quadratic using our new sum: \[x^2+3x+5x + 15.\]

Lastly, we can factor by grouping, factoring the first two terms of our expression and the last two terms: \[(x^2+3x)+(5x+15) = x(x+3) + 5(x+3) = (x+5)(x+3).\]

The factored form of \(x^2+8x+15\) is \((x+3)(x+5).\)

Factor \(x^2-4x-12.\)

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The product of \(a\) and \(c\) is \((1)(-12)=-12.\)

A factor pair of \(-12\) that sums to \(-4\) is \(-6\) and \(2.\)

Rewriting our expression, we have \(x^2-4x-12 = x^2-6x+2x-12.\)

Grouping and simplifying, we have \(x^2-6x+2x-12 = (x^2-6x)+(2x-12) = x(x-6)+2(x-6) = (x+2)(x-6).\)

The factored form of \(x^2-4x-12\) is \((x+2)(x-6).\)

Factor \(x^2-7x+6.\)

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Answer

The product of \(a\) and \(c\) is \((1)(6)=6.\)

A factor pair of \(6\) that sums to \(-7\) is \(-1\) and \(-6.\)

Rewriting our expression, we have \(x^2-7x+6 = x^2-1x-6x+6.\)

Grouping and simplifying, we have \(x^2-1x-6x+6 = (x^2-1x)+(-6x+6) = x(x-1)-6(x-1) = (x-6)(x-1).\)

The factored form of \(x^2-7x+6\) is \((x-6)(x-1).\)

If the following statement is true, what is the value of \(a\,?\)

\[x^2+5x-14=(x+a)(x-2)\]

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Answer

Observe that the left-hand side of the equation can be factorized as \[\begin{align} x^2+5x-14 &= x^2+7x-2x+7 \cdot (-2) \\ &= x(x+7)-2(x+7) \\ &= (x+7)(x-2). \end{align}\] Equating this with the right-hand side gives \(a=7\).

Remember that quadratic expressions may be written in standard form as \(ax^2+bx+c.\) Let's factor some expressions in which \(a \neq 1.\)

We'll begin with \(2x^2+7x+3,\) and follow the same steps as above..

First, we need to find the product \(ac:\) \(a\cdot c = 2\cdot 3 = 6.\)

Next, we need to find a factor pair of \(ac\) that sums to \(b.\) So we need a factor pair of 6 that sums to 7. The factor pair of 6 and 1 sums to 7.

Next, we need to rewrite the "\(b\)-term" of our quadratic using our new sum: \[2x^2+6x+1x+3.\]

Lastly, we can factor by grouping, factoring the first two terms of our expression and the last two terms: \[(2x^2+6x)+(1x+3) = 2x(x+3) + 1(x+3) = (2x+1)(x+3).\]

The factored form of \(2x^2+7x+3\) is \((2x+1)(x+3).\)

Factor \(3x^2+10x-8.\)

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Answer

The product of \(a\) and \(c\) is \((3)(-8)=-24.\)

A factor pair of \(-24\) that sums to \(10\) is \(12\) and \(-2.\)

Rewriting our expression, we have \(3x^2+10x-8 = 3x^2+12x-2x-8.\)

Grouping and simplifying, we have \(3x^2+12x-2x-8 = (3x^2+12x)+(-2x-8) = 3x(x+4)-2(x+4) = (3x-2)(x+4).\)

The factored form of \(3x^2+10x-8\) is \((3x-2)(x+4).\)

Sometimes, we can simplify a quadratic expression by factoring out a common factor before we completely factor the trinomial.

For example, let's factor \(6x^2-15x-36.\)

Every term in the trinomial is divisible by 3 so let's factor out a 3: \(6x^2-15x-36=3(2x^2-5x-12).\)

Now we can factor \(2x^2-5x-12\) following the same process that we used above.

The value of \(ac\) is \((2)(-12) = -24.\) The factor pair of \(-24\) that sums to \(-5\) is 8 and \(-3.\)

Therefore, \[3(2x^2-5x-12) = 3(2x^2+8x-3x-12) = 3[(2x^2+8x)+(-3x-12)]=3[2x(x+4)-3(x+4)]=3(2x-3)(x+4).\]

Factor \(20x^2-30x+10.\)

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Answer

Let's begin by factoring out a 10: \(20x^2-30x+10=10(2x^2-3x+1).\)

The product of \(a\) and \(c\) is \((2)(1)=2.\)

A factor pair of \(2\) that sums to \(-3\) is \(-1\) and \(-2.\)

Therefore, \[10(2x^2-3x+1)=10[2x^2-1x-2x+1]=10[(2x^2-1x)+(-2x+1)]=10[x(2x-1)-1(2x-1)]=10(x-1)(2x-1).\]

Given a quadratic equation \(ax^2+bx+c=0\), how can we factor it?

First, we need to know that the factored form of a quadratic equation is \(a(x-r_1)(x-r_2)\), where \(r_1\) and \(r_2\) are the roots of the equation and \(a\) is the coefficient of the first term.

By expanding this, we get \(a\left(x^2-(r_1+r_2)x+r_1 r_2\right)\).

Now we can try to factor the equation, but first we need to factor out \(a\), which is the coefficient of the first term. Then, we can find the roots of the equation by trying out some values, since we know \(\frac{-b}{a} = r_1+r_2\) and \(\frac{c}{a}=r_1 r_2\).

Factorize \(x^2-7x+6\).

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We don't need to factor out \(a\) since \(a=1\).

Now that we know \(r_1+r_2=-b=7\) and \(r_1 r_2=c=6\), we know that the values \(1\) and \(6\) satisfy the condition for \(r_1\) and \(r_2\).

Therefore, \[x^2-7x+6=(x-6)(x-1). \ _\square\]

Given a quadratic equation \(ax^2 + bx + c = 0\), we can factor it easily using methods described here. This, however, may not be the best method when dealing with the general quadratic with solutions which are not real. For these cases, the quadratic formula might be more applicable.

Consider the quadratic equation \(ax^2+bx+c=0\). If we divide the whole expression by \(a\), we get \[x^2 + \frac{b}{a}x + \frac{c}{a} = 0 .\] Rearranging gives \[x^2 + \frac{b}{a}x = - \frac{c}{a} .\] Let us now add \(\frac{b^2}{4a^2}\) to both sides: \[\begin{align} x^2 + \frac{b^2}{4a^2} + \frac{b}{a}x &= \frac{b^2}{4a^2} - \frac{c}{a}\\ \left(x + \frac{b}{2a}\right)^2 &= \frac{b^2-4ac}{4a^2} . \end{align}\] Taking the square root of both sides, we have \[x + \frac{b}{2a} = \pm \sqrt{\frac{b^2-4ac}{4a^2}} .\] Rearranging one last time, we get \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} .\]

Notice that now we are able to find the roots of a quadratic even if they are not real. This allows us to factor it very easily. Thus for any quadratic \(ax^2 + bx + c = 0\), we can factor it into \(k(x+\alpha)(x+\beta) = 0 \), where \(\alpha= \frac{-b+ \sqrt{b^2 - 4ac}}{2a}, \) \( \beta= \frac{-b - \sqrt{b^2 - 4ac}}{2a}, \) and \(k\) is some constant.

Factorize \(x^2 + x - 1 = 0.\)

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Using our formula, we obtain \[\phi= \frac{-b+ \sqrt{b^2 - 4ac}}{2a} = \frac{-1 +\sqrt{5}}{2}, \Phi= \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{-1 - \sqrt{5}}{2} .\] Our quadratic is thus \[k(x-\phi)(x-\Phi) = 0\] for some constant \(k\). However, since the leading coefficient of our first term is 1, we know that the factorization must be \[(x-\phi)(x - \Phi) = \left(x - \frac{-1 +\sqrt{5}}{2} \right)\left(x- \frac{-1 - \sqrt{5}}{2}\right). \ _\square\]

This method of factorization also works when we are not dealing with quadratics with real solutions/coefficients.

Factorize \(2x^2 +2 x + 2 = 0.\)

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We first observe that this has no real solutions because its discriminant is negative: \[D = b^2 - 4ac = 2^2 - 4\times2\times2 = -12 < 0.\] We can proceed to finding its roots using \[x = \frac{-b+ \sqrt{b^2 - 4ac}}{2a} =\frac{-b+ \sqrt{D}}{2a} . \] Using our formula \[\alpha= \frac{-2+ \sqrt{-12}}{2.2} =\frac{-2+ \sqrt{4\times-1\times 3}}{4} =\frac{-2+ 2i\sqrt{3}}{4} = \frac{-1}{2} + \frac{i\sqrt{3}}{2} . \] Similarly, we get \[\beta= \frac{-2 - \sqrt{-12}}{2.2} =\frac{-2 - \sqrt{4\times-1\times 3}}{4} =\frac{-2 - 2i\sqrt{3}}{4} = \frac{-1}{2} - \frac{i\sqrt{3}}{2} .\]

We can now write our equation as \[k\left(x-\frac{-1}{2} - \frac{i\sqrt{3}}{2}\right)\left(x-\frac{-1}{2} + \frac{i\sqrt{3}}{2}\right)=0.\] Notice that \(k\) has to be \(2\) in order to make the coefficient of \(x^2\) \(2\). Thus, we finally get \[2\left(x+\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)\left(x+\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)=0. \ _\square\]

This allows us to factorize quadratics with irrational and even imaginary coefficients.

Factorize \( \sqrt{24}x^2 + 5\sqrt{2}x + \sqrt{6} = 0.\)

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Again, our discriminant is \[D = b^2 - 4\times a\times c = 50 - 4\times12 = 2 > 0. \] Using our formula we get \[\begin{align} \alpha&= \frac{-5\sqrt{2} + \sqrt{2} }{4\sqrt{6}} = \frac{ - 4\sqrt{2} }{ 4\sqrt{6}} = -\frac{\sqrt{3}}{3 } \\ \beta&= \frac{-5\sqrt{2} - \sqrt{2} }{4\sqrt{6}} = \frac{ - 6\sqrt{2} }{ 4\sqrt{6}} = -\frac{\sqrt{3}}{2}. \end{align} \]

Thus, our factorization becomes

\[k\left(x+\frac{\sqrt{3}}{3 }\right)\left(x + \frac{\sqrt{3}}{2} \right) = 0. \]

Again, by comparing this with our initial polynomial, we see that \(k = \sqrt{24} = 2\sqrt{6} \). Thus, our final factorized form is \[\sqrt{24}\left(x+\frac{\sqrt{3}}{3} \right)\left(x + \frac{\sqrt{3}}{2} \right) = 0. \ _\square \]