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2.4: Adding and Subtracting Polynomial Expressions

  • Page ID
    92359
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    In this section we concentrate on adding and subtracting polynomial expressions, based on earlier work combining like terms in Ascending and Descending Powers. Let’s begin with an addition example.

    Example \(\PageIndex{1}\)

    Simplify: \[\left(a^{2}+3 a b-b^{2}\right)+\left(4 a^{2}+11 a b-9 b^{2}\right) \nonumber \]

    Solution

    Use the commutative and associative properties to change the order and regroup. Then combine like terms.

    \[\begin{aligned}\left(a^{2}+3 a b-b^{2}\right) &+\left(4 a^{2}+11 a b-9 b^{2}\right) \\ &=\left(a^{2}+4 a^{2}\right)+(3 a b+11 a b)+\left(-b^{2}-9 b^{2}\right) \\ &=5 a^{2}+14 a b-10 b^{2} \end{aligned} \nonumber \]

    You Try \(\PageIndex{1}\)

    Simplify: \(\left(3 s^{2}-2 s t+4 t^{2}\right)+\left(s^{2}+7 t s-5 t^{2}\right)\)

    Answer

    \(4 s^{2}+5 s t-t^{2}\)

    If you are comfortable skipping a step or two, it is not necessary to write down all of the steps shown in Example \(\PageIndex{1}\). Let’s try combining like terms mentally in the next example.

    Example \(\PageIndex{2}\)

    Simplify: \[\left(x^{3}-2 x^{2} y+3 x y^{2}+y^{3}\right)+\left(2 x^{3}-4 x^{2} y-8 x y^{2}+5 y^{3}\right) \nonumber \]

    Solution

    If we use the associative and commutative property to reorder and regroup, then combine like terms, we get the following result.

    \[\begin{aligned}\left(x^{3}-2 x^{2} y\right.&+3 x y^{2}+y^{3} )+\left(2 x^{3}-4 x^{2} y-8 x y^{2}+5 y^{3}\right) \\ &=\left(x^{3}+2 x^{3}\right)+\left(-2 x^{2} y-4 x^{2} y\right)+\left(3 x y^{2}-8 x y^{2}\right)+\left(y^{3}+5 y^{3}\right) \\ &=3 x^{3}-6 x^{2} y-5 x y^{2}+6 y^{3} \end{aligned} \nonumber \]

    However, if we can combine like terms mentally, eliminating the middle step, it is much more efficient to write:

    \[\begin{array}{l}{\left(x^{3}-2 x^{2} y+3 x y^{2}+y^{3}\right)+\left(2 x^{3}-4 x^{2} y-8 x y^{2}+5 y^{3}\right)} \\ {\quad \quad=3 x^{3}-6 x^{2} y-5 x y^{2}+6 y^{3}}\end{array} \nonumber \]

    You Try \(\PageIndex{2}\)

    Simplify: \(\left(-5 a^{2} b+4 a b-3 a b^{2}\right)+\left(2 a^{2} b+7 a b-a b^{2}\right)\)

    Answer

    \(-3 a^{2} b+11 a b-4 a b^{2}\)

    Negating a Polynomial

    Before attempting subtraction of polynomials, let’s first address how to negate or “take the opposite” of a polynomial. First recall that negating is equivalent to multiplying by \(−1\).

    Negating

    If \(a\) is any number, then

    \[−a =(−1)a. \nonumber \]

    That is, negating is equivalent to multiplying by \(−1\).

    We can use this property to simplify \(−(a + b)\). First, negating is identical to multiplying by \(−1\). Then we can distribute the \(−1\).

    \[\begin{aligned} -(a+b) &= (-1)(a+b) \qquad \color {Red} \text { Negating is equivalent to multiplying by } -1 \\ &= (-1) a+(-1) b \qquad \color {Red} \text { Distribute the }-1 . \\ &= -a+(-b) \qquad \color {Red} \text { Simplify: }(-1) a=-a \text { and }(-1) b=-b \\ &= -a-b \qquad \color {Red} \text { Subtraction means add the opposite. } \end{aligned} \nonumber \]

    Thus, \(−(a + b)=−a −b\). However, it is probably simpler to note that the minus sign in front of the parentheses simply changed the sign of each term inside the parentheses.

    Negating a Sum

    When negating a sum of terms, the effect of the minus sign is to change each term in the parentheses to the opposite sign. \[−(a + b)=−a−b\nonumber\]

    Let’s look at this principle in the next example.

    Example \(\PageIndex{3}\)

    Simplify: \(-\left(-3 x^{2}+4 x-8\right)\)

    Solution

    First, negating is equivalent to multiplying by \(−1\). Then distribute the \(−1\).

    \[\begin{aligned} -&\left(-3 x^{2}+4 x-8\right) \\ &=(-1)\left(-3 x^{2}+4 x-8\right) \qquad \color {Red} \text{Negating is equivalent to multiplying by } -1\\ &=(-1)\left(-3 x^{2}\right)+(-1)(4 x)-(-1)(8) \qquad \color {Red} \text {Distribute the } -1 \\ &=3 x^{2}+(-4 x)-(-8) \qquad \color {Red} \text {Simplify: } (-1)(-3 x^{2})=3 x^{2}, (-1)(4 x)=-4 x, \text {and} (-1)(8)=-8\\ &=3 x^{2}-4 x+8 \qquad \color {Red} \text {Subtraction means add the opposite.} \end{aligned} \nonumber \]

    Alternate solution:

    As we saw above, a negative sign in front of a parentheses simply changes the sign of each term inside the parentheses. So it is much more efficient to write\[−(−3x^2 +4x−8) = 3x^2 −4x +8 \nonumber \]simply changing the sign of each term inside the parentheses.

    You Try \(\PageIndex{3}\)

    Simplify: \(-\left(2 x^{2}-3 x+9\right)\)

    Answer

    \(-2 x^{2}+3 x-9\)

    Subtracting Polynomials

    Now that we know how to negate a polynomial (change the sign of each term of the polynomial), we’re ready to subtract polynomials.

    Example \(\PageIndex{4}\)

    Simplify: \(\left(y^{3}-3 y^{2} z+4 y z^{2}+z^{3}\right)-\left(2 y^{3}-8 y^{2} z+2 y z^{2}-8 z^{3}\right)\)

    Solution

    First, distribute the minus sign, changing the sign of each term of the second polynomial.

    \[\left(y^{3}-3 y^{2} z+4 y z^{2}+z^{3}\right)-\left(2 y^{3}-8 y^{2} z+2 y z^{2}-8 z^{3}\right)=y^{3}-3 y^{2} z+4 y z^{2}+z^{3}-2 y^{3}+8 y^{2} z-2 y z^{2}+8 z^{3} \nonumber \]

    Regroup, combining like terms. You may perform this next step mentally if you wish.

    \[\begin{array}{l}{=\left(y^{3}-2 y^{3}\right)+\left(-3 y^{2} z+8 y^{2} z\right)+\left(4 y z^{2}-2 y z^{2}\right)+\left(z^{3}+8 z^{3}\right)} \\ {=-y^{3}+5 y^{2} z+2 y z^{2}+9 z^{3}}\end{array} \nonumber \]

    You Try \(\PageIndex{4}\)

    Simplify: \(\left(4 a^{2} b+2 a b-7 a b^{2}\right)-\left(2 a^{2} b-a b-5 a b^{2}\right)\)

    Answer

    \(2 a^{2} b+3 a b-2 a b^{2}\)

    Some Applications

    Recall that the area of a rectangle having length \(L\) and width \(W\) is found using the formula \(A = LW\). The area of a square having side s is found using the formula \(A = s^2\) (see Figure \(\PageIndex{1}\)).

    fig 5.4.1.png
    Figure \(\PageIndex{1}\): Area formulae for the rectangle and square.
    Example \(\PageIndex{5}\)

    Find the area of the square in Figure \(\PageIndex{2}\) by summing the area of its parts.

    fig 5.4.2.png
    Figure \(\PageIndex{2}\): Find the sum of the parts.

    Solution

    Let’s separate each of the four pieces and label each with its area (see Figure \(\PageIndex{3}\)).

    fig 5.4.3.png
    Figure \(\PageIndex{3}\): Finding the area of each of the four parts.

    The two shaded squares in Figure \(\PageIndex{3}\) have areas \(A_1 = x^2\) and \(A_3 = 9\), respectively. The two unshaded rectangles in Figure \(\PageIndex{3}\) have areas \(A_2 =3 x\) and \(A_4 =3x\). Summing these four areas gives us the area of the entire figure.

    \[\begin{aligned} A &=A_{1}+A_{2}+A_{3}+A_{4} \\ &=x^{2}+3 x+9+3 x \\ &=x^{2}+6 x+9 \end{aligned} \nonumber \]

    You Try \(\PageIndex{5}\)

    Find the area of the square shown below by summing the area of its parts.

    Ex 5.4.7.png
    Figure \(\PageIndex{4}\)
    Answer

    \(x^{2}+8 x+16\)


    This page titled 2.4: Adding and Subtracting Polynomial Expressions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Katherine Skelton.