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Polynomials

A polynomial has one or more terms with each term can be a constant, or a variable, or a product of constants and/or variables consisting only of positive integer exponents variables.

Examples:

$-8$

$-4x^3y$

$4x-3x^2$

$3x^2y+4x-7xy$

A polynomial of one term is called a monomial. In the above examples, the first two expressions are monomial.

A polynomial of two terms is called a binomial. The third expression in the above examples is a binomial.

A polynomial of three terms is called a trinomial. The last expression in the examples is a trinomial.

The sum of the exponents of the variables of a term is called the degree of the term. The highest degree of the terms of a polynomial is called the degree of the polynomial.

In the examples above, the first one, $-8$ has a degree of $0$ as there is no variable.

The degree of the second polynomial, $-4x^3y$ is $4$ as the sum of the exponents of the variable is $3+1=4$.

The polynomial, $4x-3x^2$ has two terms, with the degree of the first term is $1$ and that of the second one is $2$. The highest is $2$, so $2$ is the degree of the polynomial.

The polynomial, $3x^2y+4x-7xy$ has three terms, with the degree of first term is $3$ and that of the second and third terms are $1$ and $2$ respectively. So the highest degree is $3$, which is the degree of the polynomial.

Integer exponents

Properties of exponents

The product rule:multiplication of like bases

$a^m\cdot a^n=a^{m+n}$

The quotient rule:division of like bases

$\dfrac{a^m}{a^n}=a^{m-n}$

The power rule:

$(a^m)^n=a^{mn}$

Power of a product:

$(ab)^m=a^m\cdot b^m$

Power of a quotient:

$\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}$

The zero exponent:

$a^0=1$

Proof:

We can write

$a^0=a^{n-n}$

Applying the quotient rule on the right hand side,

$a^0=\dfrac{a^n}{a^{n}}=1$

The negative exponent:

$a^{-n}=\dfrac{1}{a^n}$

By using the properties of exponents we can simplify expressions with exponents.

Example 1:Simplify, $\left (8a^5b^4 \right )^3$.

Solution: Applying the power of a product rule and the power rule,
$\begin{align*} \left (8a^5b^4 \right )^3 &= 8^3(a^5)^3(b^4)^3\\ &= 8^3a^{5\cdot 3}b^{4\cdot 3}\\ &=512a^{15}b^{12}\\ \end{align*} $
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Example 2: Simplify, $\left (8a^5b^4 \right )^{-4}$.

solution:

Applying the rule of negative exponent,
$\begin{align*} \left (2b^3c^4 \right )^{-4} &= \dfrac{1}{ \left (2b^3c^4 \right )^{4}} \\ &=\dfrac{1}{8\,b^{12}\,c^{16}}\\ \end{align*}\\$
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Example 3:Simplify, $(2x^3yb^2)^3(5x^2y^5)$

Solution:

$\begin{align*} \left(2x^3yb^2\right)^3 \left (5x^2y^5\right) &= \left(8x^9y^3b^6\right)\left (5x^2y^5\right)\\ &=40x^{11}y^8b^6 \\ \end{align*}\\$
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Example 4:Simplify, $\left(\dfrac{4u^8b^7z}{3u^3b^6z^3}\right)^2$

Solution:

First simplify what is within the parenthesis. Then apply the power rule.
$\begin{align*} \left(\dfrac{4u^8b^7z}{3u^3b^6z^3}\right)^2 &= \left(\dfrac{4u^5b}{3z^2}\right)^2 &=\dfrac{16u^{10}b^2}{9z^4} \end{align*}\\$
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Example 5:Simplify $\left(\dfrac{4u^8b^3\,z^3}{3u^3b^6z}\right)^{-3}$

Solution:
Take the reciprocal of the expression within the parenthesis, then apply the power rule.
$\begin{align*} \left(\dfrac{4u^8b^3z^3}{3u^3b^6z}\right)^{-3} &= \left(\dfrac{3u^3b^6z}{4u^8b^3\,z^3}\right)^{3}\\ &= \left(\dfrac{3\,b^3}{4u^5\,z^2}\right)^{3}\\ &= \dfrac{27\,b^9}{64\,u^{15}\,z^6} \end{align*}\\$
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Example 6:Simplify $8\,z^{-7}\,y^2\cdot 5\,z^3\,a^2\,y^{-5}$

Solution:

$\begin{align*} 8\,z^{-7}\,y^2\cdot 5\,z^3\,a^2\,y^{-5} &= 8\cdot 5 \,z^{-7+3}\,y^{2-5}a^2\\ &=40\,z^{-4}y^{-3}a^2\\ &=\dfrac{40\,a^2}{z^4\,y^3} \end{align*}\\$
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Example 7:Simplify $\dfrac{7x^{-8}\,y^{11}}{b^5\,x^5\,y^4}$

Solution:

$\begin{align*} \dfrac{7x^{-8}\,y^{11}}{b^5\,x^5\,y^4} &= \dfrac{7\,y^{11-4}}{b^5\,x^{5+8}}\\ &=\dfrac{7\,y^7}{b^5\,x^{13}}\\ \end{align*}\\$
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Scientific notation

Scientific notation is a convenient way to write a very small or a very large number. It is written in the form,

$a\times 10^n$, where $n$ is a positive or a negative integer and $1\le a \lt 10$.

Example 1: Write in scientific notation, $7600000000$

Answer: The number is greater than 10, and has no decimal point. So, put a decimal point after the first digit, $7$. Now, count and find howmany digits are there after the decimal point, there are $9$ digits. This number is the $n$. So, $n=9$.

Therefore, $7600000000=7.600000000\times 10^9$

Ignore all the zeros after the last non-zero digit, $6$, as they have no values. You get,

$7600000000=7.6\times 10^9$

Example 2: Write in scientific notation, $0.000000065$

Answer:The number is less than $1$. So, move the decimal point to the right and put it after the first non-zero digit, $6$. Since you moved the decimal point to the right, $n$ will be negative. Now, count the number of places the decimal point has moved, which is $8$. Negative of this number is the $n$. So, $n=-8$.

Therefore, $0.000000065=6.5\times 10^{-8}$

Note that I ignored the zeros before the digit $6$ as they have no values.

Example 3: Write the following number in standard notation, $4.82\times 10^{-6}$

Answer: Since the exponent is negative, you need to move the decimal point to the left. The exponent is $-6$, so move the point 6 places to the left. Since there is only one digit in the left of the decimal point, add 5 zeros for the remaining places.

So, $4.82\times 10^{-6}=0.00000482$

Note that I have put a zero on the left of the decimal point, which is optional.

Example 4: Write $6.25\times 10^5$ in standard notation.

Answer:Since the exponent is positive, move the decimal point five places to the right, add zeros if there is not enough places. Remove the point, if there is no non-zero digits after the decimal point.

So, $6.25\times 10^5=625000$

Addition and subtraction of polynomials and polynomial function

When adding/subtracting two or more polynomials, you need to combine the like terms.

Like terms are the terms that have the same variables and the variables are raised to the same powers. i.e., Like terms are terms they differ only by a constant factor.

Examples of like terms: $8x^2$ and $-4x^2$; $-2xy$ and $11xy$; $3xy^3$ and $xy^3$


Example:1 Simplify, $(-2x^2+3y-y^2)-(4x^2+2y-5y^2)$

Solution

First remove the parentheses and then combine the like terms ().:
$\begin{align*} (-2x^2+3y-y^2)-(4x^2+2y-5y^2)&=-2x^2+3y-y^2-4x^2-2y+5y^2\\ &=-6x^2+y+4y^2 \end{align*}$
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Example:2 Simplify, $(3a^2b-4ab+4ab^2)-(5a^2b-8ab+ab^2)$

Solution:

$\begin{align*} (3a^2b-4ab+4ab^2)-(5a^2b-8ab+ab^2)&=3a^2b-4ab+4ab^2-5a^2b+8ab-ab^2\\ &=-2a^2b+4ab+3ab^2 \end{align*}$
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Multiplication of polynomials

When multiplying two polynomials, we use the distributive formula:

$a(b+c)=ab+ac$

Example 1:Multiply and simplify, $(4x-2)(3x+5)$

Solution:

$\begin{align*} (4x-2)(3x+5)&=4x\cdot 3x+4x\cdot 5-2\cdot 3x-2\cdot5\\ &= 12x^2+20x-6x-10\\ &=12x^2+14x-10 \end{align*}$
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Example 2:Write without parenthesis and simplify, $(4x-3u)^2$

Solution:

$\begin{align*} (4x-3u)^2 &=(4x-3u)(4x-3u)\\ &= 16x^2-12xu-12xu+9u^2\\ &=16x^2-24xu+9u^2 \end{align*}$

Example 2:Write without parenthesis and simplify, $(4x-3u)^2$

Solution:

$\begin{align*} (4x-3u)^2 &=(4x-3u)(4x-3u)\\ &= 16x^2-12xu-12xu+9u^2\\ &=16x^2-24xu+9u^2 \end{align*}$
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Example 3:Multiply and simplify, $(-4y-3)(5y^2-2y+5)$

Solution:

$\begin{align*} (-4y-3)(5y^2-2y+5) &=-20y^3+8y^2-20y-15y^2+6y-15\\ &= -20y^3-7y^2-14y-15 \end{align*}$
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Division of polynomials

When dividing a polynomial by a monomial, you need to divide each term of the polynomial by the monomial,

$\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$

Example:1 Divide, $\dfrac{24x^4y-8x^3y+12x^2y^5}{4x^2y}$. Simply the answer as much as possible.

Solution:

$\begin{align*} \dfrac{24x^4y-8x^3y+12x^2y^5}{4x^2y} &= \dfrac{24x^4y}{4x^2y}-\dfrac{8x^3y}{4x^2y}+\dfrac{12x^2y^5}{4x^2y} \\ &= 6x^2-2x+3y^4 \end{align*}$

Long division

Example 1: Divide, $12x^3+29x^2+11x-30 \div 4x+3$

Solution:

We have the dividend:

$12x^3+29x^2+11x-30 $

and the divisor:

$4x+3$

1. you should arrange the dividend and the divisor in the descending order, i.e., in the descending power of the variable.

$12x^3+29x^2+11x-30 $

2. If a term is missing, for example, you have $x^3$ term, and an $x$ term, but no $x^2$ term, then add an $x^2$ term with zero coefficient between the $x^3$ and the $x$ term.

Next, start the division,

$4x+3\overline{)12x^3+29x^2+11x-30}$

Now, take the leading term of the dividend and divide that by the leading term of the divisor.

$\dfrac{12x^3}{4x}=3x^2$. Put this ratio as quotient.

$ \hspace{15.5mm}3x^2$
$ 4x+3\overline{)12x^3+29x^2+11x-30}$

Next, multiply the ratio with the divisor:

$3x^2(4x+3)=12x^3+9x^2$. Write this below and subtract,

$ \hspace{15.5mm}3x^2$
$ 4x+3\overline{)12x^3+29x^2+11x-30}$
$\hspace{16mm}\underline{12x^3+9x^2}$
$\hspace{31.5mm}20x^2$

Bring the next term (the $x$ term) down

$ \hspace{15.5mm}3x^2$
$ 4x+3\overline{)12x^3+29x^2+11x-30}$
$\hspace{16mm}\underline{12x^3+9x^2}$
$\hspace{31.5mm}20x^2+11x$

Take the ratio of the leading term of the last expression with the leading term of the divisor.

$\dfrac{20x^2}{4x}=5x$. Add this in the quotient.

$ \hspace{15.5mm}3x^2+5x$
$ 4x+3\overline{)12x^3+29x^2+11x-30}$
$\hspace{16mm}\underline{12x^3+9x^2}$
$\hspace{31.5mm}20x^2+11x$

Multiply the ratio with the divisor:

$5x(4x+3)=20x^2+15x$. Now subtract this,

$ \hspace{15.5mm}3x^2+5x$
$ 4x+3\overline{)12x^3+29x^2+11x-30}$
$\hspace{16mm}\underline{12x^3+9x^2}$
$\hspace{31.5mm}20x^2+11x$
$\hspace{31.5mm}\underline{20x^2+15x}\\$
$\hspace{43.5mm}-4x$

Bring the next term, $-30$ down

$ \hspace{15.5mm}3x^2+5x$
$ 4x+3\overline{)12x^3+29x^2+11x-30}$
$\hspace{16mm}\underline{12x^3+9x^2}$
$\hspace{31.5mm}20x^2+11x$
$\hspace{31.5mm}\underline{20x^2+15x}\\$
$ \hspace{43.5mm}-4x-30$

Divide the leading term by the leading term of the divisor and add it to the quotient.

$\dfrac{4x}{-4x}=-1$

Then, multiply this ratio with the divisor and subtract as before.

$-1(4x+3)=-4x-3$,

$ \hspace{15.5mm}3x^2+5x-1$
$ 4x+3\overline{)12x^3+29x^2+11x-30}$
$\hspace{16mm}\underline{12x^3+9x^2}$
$\hspace{31.5mm}20x^2+11x$
$\hspace{31.5mm}\underline{20x^2+15x}\\$
$ \hspace{43.5mm}-4x-30$
$ \hspace{45.5mm}\underline{-4x-3}\\$
$ \hspace{55.5mm}-27$
Now, write the answer,

Quotient: $3x^2+5x-1$

Remainder: $-27$

Factoring polynomial

Writing an integer as a product of two or more integers is called factoring. For example, if we write $18$ as $2\cdot 9$, we say that we factored the number $18$. The two numbers $2$ and $9$ are called the factors. $18$ can also be written as $3\cdot6$, so now $3$ and $6$ are the factors of $18$.

Like integers, we can factor out polynomials, i.e., we can write a polynomial as a product of two or more polynomials. For example, we can write

$8x^3+4x=4x(2x+1)$. Here, the a binomial is written as a product of a monomial and a binomial.

When factoring out a polynomial, first we need to find the greatest common factor.

Greatest common factor (GCF) is the greatest possible factor that can divide each terms of a polynomial.

Example 1: Factor, $6c^2+8c^5$

Solution:

There are two terms in this problem, we need to find the greatest common factor for the constants and that for the variables of the terms.

For the constants $6$ and $8$, the GCF is $2$

For the variable, $c$, the GCF is $c^2$ as $2$ is the smallest exponent of the variable $c$ among the two terms.

Now, we can write,

$\begin{align*} 6c^2+8c^5 &=2\cdot 3 c^2+2\cdot4c^2c^3\\ &=2c^2(3+4c^3) \end{align*}$

Example 2:Factor, $12x^3y^2-48xy^5v^3$

Solution:

For the constants, GCF$=12$
For the variables, take the least power exponents of $x$ and $y$. Since the variable $v$ is in only one of the terms, it has no common factor. So, GCF$=xy^2.$

$\begin{align*} 12x^3y^2-48xy^5v^3 &= 12x\cdot x^2y^2-12\cdot 4xy^2\cdot y^3v^3\\ &=12xy^2(x^2-4y^3v^3) \end{align*}$

Factoring by grouping

Sometimes, we can factor a polynomial by taking the polynomial as groups and factor each group separately, called factor by grouping.

Example 1:Factor, $5x^3-2x^2-15x+6$

Solution:Consider the polynomial as two groups, factor them each and finally factor out the common factors from the groups.

There are four terms, we take two as one group.

$\begin{align*} 5x^3-2x^2-15x+6 &= \underline{5x^3-2x^2}\:\:\underline{-15x+6}\\ &= x^2(5x-2)-3(5x-2)\\ &= x^2(5x-2)-3(5x-2)\\ &= (5x-2)(x^2-3) \end{align*}$

Example 2:Factor, $3y^4-5y^3+6y-10$

Solution:Group the terms 2 by 2, factor each group and finally factor out the common factors from the groups.

$\begin{align*} 3y^4-5y^3+6y-10 &= \underline{3y^4-5y^3}\:\:\underline{+6y-10}\\ 3y^4-5y^3+6y-10 &= y^3(3y-5)+2(3y-5)\\ &= (3y-5)(y^3+2) \end{align*}$

Factoring a trinomials

We have a trinomial, $ax^2+bx+c$

To solve this first,

1. Multiply the coefficient of $x^2$, i.e., $a$ and the constant term, $c$, take that as $p$, then

2. Now you need to find two numbers split the coefficient of $x$ into two numbers so that the sum of the two numbers is equal to $b$, the coefficient of $x$ and their product is equal to $p$.

Example:1Factor, $x^2-8x+15$

Solution:

Multiply the coefficient of $x^2$ and the constant term we get $1\cdot15=15$

Find the factors of $15$,

$(1,15),(3,5),(-1,-15),(-3,-5)$. Out of these pick one that if you add, you should get $-8$.

That is $-3$ and $-5$

Sow, we can write,

$\begin{align*} x^2-8x+15 &= x^2-5x-3x+15 \\ &= x(x-5)-3(x-5) &= (x-5)(x-3) \end{align*}$

Example 2:Solve, $6x^2-13x-5$

Solution:

Multiply, the coefficient of $x^2$ and the constant, i.e., $6\cdot -5=-30$

Find the factors of $-30$. Since the number is negative, one of the factor should be negative:

$(1,-30),(2,-15),(3,-10),(5,-6)$ and $(-1,30),(-2,15),(-3,10),(-5,6)$

Now, pick one of these that when we add should give you the coefficient of $x$, $-13$ that is $(2,-15)$

So, now we can write,

$\begin{align*} 6x^2-13x-5 &= 6x^2+2x-15x-5 \\ &= 2x(3x+1)-5(3x+1)\\ &= (3x+1)(2x-5) \end{align*}$

Example:Factor, $24y^2-34y+12$

Solution:

In the above equation, you see that there is a common constant factor, $2$. So, first factor that out. Not all trinomial has a constant common factor, but if there is a common factor, first you need to take that out. So, we have,

$24y^2-34y+12 = 2(12y^2-17y+6)$

Now, factor the trinomial that is inside the parentheses,

Multiply the coefficient of $y^2$ with the constant term, i.e., $12\times 6=72$

Find the factors for $72$:

$(1,72),(2,36),(3,24),(4,18),(6,12),(8,9)$ and $(-1,-72),(-2,-36),(-3,-24),(-4,-18),(-6,-12),(-8,-9)$

Now, pick one that if you add should give you, $-17$. That is $(-8,-9)$

Now,

$\begin{align*} 24y^2-34y+12 &= 2(12y^2-8y-9y+6)\\ &= 2(4y(3y-2)-3(3y-2)) \\ &=2(3y-2)(4y-3) \end{align*}$

Example:Factor, $-6x^2-7x+5$

Solution:

It is important that the coefficient of the $x^2$ should be positive. So, factor out the minus sign and put a parentheses.

That is

$-6x^2-7x+5$=$-(6x^2+7x-5)$

Now, factor the trinomial within the parentheses.

Multiplying, $6$ and $-5$, we get $-30$, and factoring this, $(1,-30), (2,-15),(3,-10),(5,-6)$ and $(-1,30), (-2,15),(-3,10),(-5,6)$

The two numbers which add to give $7$ is $(-3,10)$. Therefore,

$\begin{align*} -6x^2-7x+5 &= -(6x^2-3x+10x-5)\\ &= -(3x(2x-1)+5(2x-1)) \\ &=-(2x-1)(3x+5) \end{align*}$

Factoring a perfect square trinomial

If we have a trinomial of the form, $a^2+2ab+b^2$, we can write this as a perfect square:

$(a+b)^2$.

And a trinomial of the form, $a^2-2ab+b^2$ as

$(a-b)^2$.

Example 1:factor, $4y^2+20y+25$

Solution:

The $y^2$ term, $4y^2$, is a perfect square, so taking its square root, we get $2y$. Also, the constant term $(25)$, is a perfect square, its square root is $5$.

Now, take, $2$ times the product of these square roots, $2\cdot2y\cdot5=10y$. Now compare this with the absolute value of the middle term, the $y$ term, we see that both are the same. So, now we can write the trinomial as a perfect square:

$4y^2+20y+25 = (2y)^2+2\cdot 2y\cdot 5 +5^2$
The right hand side is in the form $a^2+2ab+b^2$ with $a=2y$ and $b=5$, so we can write it as $(a+b)^2$.

Therefore,

$4y^2+20y+25 = (2y+5)^2$

Example 1:Factor, $9x^2-24xy+16y^2$

Solution:

The quadratic trinomial has two variables, $x$ and $y$. We can factor this trinomial as same as the quadaratic trinomial of one varable.

The first term is a perfect square, with square root, $3x$, and the third term is also a perfect square, with square root, $4y$. now, let us check, $2$ times of the product of these square roots is the absolute value of the middle term. $2\cdot3x\cdot4y=24xy$, this is equal to the absolute value of the middle term. So, we can write the polynomial as a perfect square.

The trinomial is now in the form $a^2-2ab+b^2$, with $a=3x$ and $b=4y$. Therefore, we can write,it as $(a-b)^2$.

i.e., $9x^2-24xy+16y^2=(3x-4y)^2$

Difference of squares

$a^2-b^2=(a+b)(a-b)$.

By using this formula, we can factor binomials of in this form.

Example:1 Factor $49-25y^2$


Solution:
We can write, both the terms a perfect square, and apply the difference of squares formula:
\begin{align*} $49-25y^2 &=7^2-(5y)^2$\\ &=(7+5y)(7-5y) \end{align*}

Example 2:Factor completely $18y^4-162y^2$.

Solution:
In the polynomial, there are common factors that we can factor out. The greatest common factor is $18y^2$. Factoring that out,

$18y^4-162y^2=18y^2(y^2-9)$

First term is already a perfect square, and we can write the second term also as a perfect square. Then applying the formula,

$\begin{align*} 18y^4-162y^2 &=18y^2(y^2-3^2)\\ &=18(y+3)(y-3) \end{align*}$

Example 3:Factor completely $18y^4-162y^2$.

Solution: Greatest common factor is $2w^4$. factoring this out

$18y^4-162y^2=2w^4(x^4-16)$.
Now, we can write $16$ as $2^4$.

$18y^4-162y^2=2w^4(x^4-2^4)$.

We can write, $x^4=(x^2)^2$ and $2^4$ as $(2^2)^2$. Therefore,

$18y^4-162y^2=2w^4((x^2)^2-(2^2)^2$.

Applying the difference of squares formula,

$18y^4-162y^2=2w^4(x^2+2^2)(x^2-2^2)$

Again we can apply the same formula to the last factor, $(x^2-2^2)$ and we will get,

$18y^4-162y^2=2w^4(x^2+4)(x+2)(x-2)$

Sum and difference of difference of cubes

Following formulas are called sum of difference of cubes,

$a^3+b^3=(a+b)(a^2-ab+b^2)$ and

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Example :Factor $64-27u^3$.


Solution:

We can write $64$ as $4^3$ and also the second term as $(3u)^3$. Then, we can apply the difference of cubes formula:



$\begin{align*} 64-27u^3 &=4^3-(3u)^3\\ &=(4-3u)(4^2+4\cdot3u+(3u)^2)\\ &=(4-3u)(16+12u+9u^2)\\ \end{align*}$

Quadratic equation

An equation of the form $ax^2+bx+c=0$, where $a\ne 0$ is called a quadratic equation. Solving a quadratic equation means find the value (s) of the variable, $x$. To solve first we will factor the polynomial on the left hand side. Then, we will apply the zero product rule.

The zero product rule is

If $ab=0$, then $a=0$ or $b=0$.

Example:1 Solve $x^2-16x+28=0$

Solution: Multiplying the coefficient of $x^2$ and the constant term, $1\cdot 28 = 28$. Now we need to find two factors that if we multiply should give 28 and if we add we should get $-16$. The two factors are $-2,-14$. Now, you replace, $-16x$ with $-2x-14x$:

$x^2-2x-14x+28=0$

Now, factor by grouping, factor the first two terms and next the last two terms:

$x(x-2)-14(x-2)=0$

Taking the common factors out,

$(x-2)(x-14)=0$

Applying the zero product rule,

$x-2=0$ or $x-14=0$

Solving for $x$,

$x=2$ or $x=14$

Example: 2 Solve $8w^2-32w=0$

Solution: In this equation there is no constant term, so just factor out the greatest common factor and apply zero product rule:
$\begin{align*} 8w^2-32w &=0\\ 8w(w-4) &=0\\ \end{align*}$

Applying the zero product rule,

$w=0$ or $w-4=0$.

or

$\boxed{w=0 \: or\: w=4}$

Example: 3 Solve $4u^2+12u=-9$

Solution: This equation is not in the $ax^2+bx+c=0$ form, so first you need to write the equation in this form. So, move the number on the right hand side to the left by adding a $9$,

$4u^2+12u+9=0$

The first and the last term can be written as perfect square,

$(2u)^2+12u+3^2=0$

This looks, of the form: $a^2+2ab+b^2$, to make sure, comparing this with our equation, we have $a=2u$ and $b=3$. Now taking the product, $2ab$, we get $2\cdot2u\cdot 3=12u$, which is same as the middle term, therefore the quadratic equation is exactly of the form, $a^2+2ab+b^2$. So we can write it as $(a+b)^2$. Therefore, we can write,

$(2u+3)^2=0$

or

$(2u+3)(2u+3)=0$

Applying the zero product rule,

$2u+3=0$ or $2u+3=0$

Since both of the above equations are the same, take one of them and solve for $x$, you will get,

$\boxed{u=-3/2}$

$\left(\dfrac{a}{b}\right )^{-n}=\left(\dfrac{b}{a}\right )^{n}$

$\left(\dfrac{a}{b}\right )^{-n}=\dfrac{a^{-n}}{b^{-n}}$