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Exercises 1 To write the given polynomial in standard form, recall that the standard form of a polynomial arranges the terms by degree in descending numerical order. -8q3​ Since the given polynomial consists of only one term, this is its standard form. We also want to classify this polynomial by degree, leading coefficient, and the number of terms. -8q3​ The highest exponent, and therefore the degree of the polynomial, is 3 and the leading coefficient of the polynomial is equal to -8. This polynomial consists of only one term, so it is a monomial. Therefore, it is a cubic monomial.
Exercises 2 To write the given polynomial in standard form, recall that the standard form of a polynomial arranges the terms by degree in descending numerical order. 9+d2−3d⇔d2−3d+9​ We want to classify this polynomial by degree, leading coefficient, and the number of terms. 1d2−3d+9​ The highest exponent of the polynomial, and therefore the degree of the polynomial, is 2 while the leading coefficient is equal to 1. This polynomial consists of three terms, so it is a trinomial. Therefore, it is a quadratic trinomial.
Exercises 3 To write the given polynomial in standard form, recall that the standard form of a polynomial arranges the terms by degree in descending numerical order. 32​m4−65​m6⇔-65​m6+32​m4​ We want to classify this polynomial by degree, leading coefficient, and the number of terms. -65​m6+32​m4​ The highest exponent of the polynomial, and therefore the degree of the polynomial, is 6 while the leading coefficient is equal to -65​. This polynomial consists of two terms, so it is a binomial. Therefore, it is a 6th-degree binomial.
Exercises 4 To write the given polynomial in standard form, recall that the standard form of a polynomial arranges the terms by degree in descending numerical order. -1.3z+3z4+7.4z2⇔3z4+7.4z2−1.3z​ We want to classify this polynomial by degree, leading coefficient, and the number of terms. 3z4+7.4z2−1.3z​ The highest exponent of the polynomial, and therefore the degree of the polynomial, is 4 while the leading coefficient is equal to 3. This polynomial consists of three terms, so it is a trinomial. Therefore, it is a quartic trinomial.
Exercises 5 The first step in simplifying this expression is to remove the parentheses. (2x2+5)+(-x2+4)⇔2x2+5−x2+4​ As a next step, we will identify which, if any, terms can be combined. Remember, only like terms, constant terms, or terms with the same variable and the same exponent can be combined. 2x2+5−x2+4​ In this case, we have two x2-terms, and two constants. The x2-terms and constants can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then evaluate the difference between the like terms. 2x2+5−x2+4Commutative Property of Addition2x2−x2+5+4Add and subtract termsx2+9
Exercises 6 The first step in simplifying this expression is to remove the parentheses. We will do that by applying the Distributive Property. (-3n2+n)−(2n2−7)Remove parentheses-3n2+n−(2n2−7)Distribute -1-3n2+n−2n2+7 As a next step, we will identify which, if any, terms can be combined. Remember, only like terms, constant terms, or terms with the same variable and the same exponent can be combined. -3n2+n−2n2+7​ In this case, we have two n2-terms, an n-term, and a constant. The n2-terms can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then evaluate the sum of like terms. -3n2+n−2n2+7Commutative Property of Addition-3n2−2n2+n+7Add terms-5n2+n+7
Exercises 7 The first step in simplifying this expression is to remove the parentheses. We will do that by applying the Distributive Property. (-p2+4p)−(p2−3p+15)Remove parentheses-p2+4p−(p2−3p+15)Distribute -1-p2+4p−p2+3p−15 As a next step, we will identify which, if any, terms can be combined. Remember, only like terms, constant terms, or terms with the same variable and the same exponent can be combined. -p2+4p−p2+3p−15​ In this case, we have two p2-terms, two p-terms, and a constant. The p2-terms and p-terms can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then evaluate the sum of like terms. -p2+4p−p2+3p−15Commutative Property of Addition-p2−p2+4p+3p−15Add terms-2p2+7p−15
Exercises 8 The first step in simplifying this expression is to remove the parentheses. (a2−3ab+b2)+(-a2+ab+b2)Remove parenthesesa2−3ab+b2−a2+ab+b2 As a next step, we will identify which, if any, terms can be combined. Remember, only like terms, constant terms, or terms with the same variable and the same exponent can be combined. a2−3ab+b2−a2+ab+b2​ In this case, we have two a2-terms, b2-terms, and two ab-terms. All of them can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then evaluate the sum and difference between the like terms. a2−3ab+b2−a2+ab+b2Commutative Property of Additiona2−a2−3ab+ab+b2+b2Add and subtract terms-2ab+2b2
Exercises 9 We want to simplify the expression by multiplying the binomials. To do so, we will apply the Distributive Property. (w+6)(w+7)Distribute (w+7)w(w+7)+6(w+7)Distribute ww2+7w+6(w+7)Distribute 6w2+7w+6w+42Add and subtract termsw2+13w+42
Exercises 10 We want to simplify the expression by multiplying the binomials. To do so, we will apply the Distributive Property. (3−4d)(2d−5)Distribute (2d−5)3(2d−5)−4d(2d−5)Distribute 36d−15−4d(2d−5)Distribute 4d6d−15−(8d2−20d)Remove parentheses and change signs6d−15−8d2+20dAdd terms26d−15−8d2
Exercises 11 To simplify the expression by multiplying a binomial by a trinomial, we will apply the Distributive Property. (y+9)(y2+2y−3)Distribute (y2+2y−3)y(y2+2y−3)+9(y2+2y−3)Distribute yy3+2y2−3y+9(y2+2y−3)Distribute 9y3+2y2−3y+9y2+18y−27Add and subtract termsy3+11y2+15y−27
Exercises 12 Notice that the given expression is a multiplication of conjugates. This allows us to use a shorter method of multiplying the terms than the normal distribution process. (3z−5)(3z+5)Commutative Property of Multiplication(3z+5)(3z−5)(a+b)(a−b)=a2−b2(3z)2−52(ab)m=ambm32z2−52Calculate power9z2−25
Exercises 13 We can find the product of this expression using polynomial identities. In this case, we have the square of a binomial. (t+5)2(a+b)2=a2+2ab+b2t2+10t+52Calculate powert2+10t+25
Exercises 14 We can find the product of this expression using polynomial identities. In this case, we have the square of a binomial. (2q−6)2(a−b)2=a2−2ab+b2(2q)2−24q+62Calculate power4q2−24q+36
Exercises 15 In order to solve the equation, we want to rewrite the left-hand side as a product of terms and then apply the Zero Product Property.Factoring We will begin by factoring out the greatest common factor (GCF) of the terms. To do so, we will consider coefficients and variables separately. 5x2−15x​ Let's start by finding the GCF of 5 and 15. Factors of 5:Factors of 15:​ 1,5 1,3,5 and 15​ We found that the GCF of the coefficients is 5. To find the GCF of the variables, we need to identify the variables repeated in both terms, and write them with their minimum exponents. Factors of 1st variable:Factors of 2nd variable:​ x x,x2​ We see that there is one repeated variable factor, x. Thus, the GCF of the expression is 5⋅x=5x. Now we can write the given expression in terms of the GCF. 5x2−15x⇔5x⋅x−x⋅3​ Finally, we will factor out the GCF. 5x⋅x−5x⋅3⇔5x(x−3)​Solving the equation Now, we will move on to solving the equation by applying the Zero Product Property. 5x(x−3)=0Use the Zero Product Property5x=0x−3=0​(I)(II)​(I):  LHS/5=RHS/5x=0x−3=0​(II):  LHS+3=RHS+3x=0x=3​ We found that there are two solutions to the equation, which are x=0 and x=3.
Exercises 16 We are going to solve the equation by applying the Zero Product Property. (8−g)(8−g)=0Use the Zero Product Property8−g=08−g=0​(I)(II)​(I), (II):  LHS+g=RHS+g8=g8=g​(I), (II):  Rearrange equationg=8g=8​ We found that the solution to the given equation is g=8.
Exercises 17 Since the left-hand side is already written in factored form, we will solve the equation by applying the Zero Product Property. (3p+7)(3p−7)(p+8)=0Use the Zero Product Property3p+7=03p−7=0p+8=0​(I)(II)(III)​(I):  LHS−7=RHS−73p=-73p−7=0p+8=0​(I):  LHS/3=RHS/3p=-37​3p−7=0p+8=0​(II):  LHS+7=RHS+7p=-37​3p=7p+8=0​(II):  LHS/3=RHS/3p=-37​p=37​p+8=0​(III):  LHS−8=RHS−8p=-37​p=37​p=-8​ We found that the solutions to the equation are: p=-37​,p=37​ and p=-8.
Exercises 18 Since the left-hand side of the equation is already written in factored form, we are going to apply the Zero Product Property. -3y(y−8)(2y+1)=0LHS/(-3)=RHS/(-3)y(y−8)(2y+1)=0Use the Zero Product Propertyy=0y−8=02y+1=0​(I)(II)(III)​(II):  LHS+8=RHS+8y=0y=82y+1=0​(III):  LHS−1=RHS−1y=0y=82y=-1​(III):  LHS/2=RHS/2y=0y=8y=-21​​ We found that the solutions to the equation are 0, 8, and -21​.
Exercises 19
Exercises 20
Exercises 21 To find the width of the bunker at ground level, we will use the x-coordinates of the points where the curve cross the x-axis.At ground level, y=0. Let's substitute it. y=-2165​(x−72)(x+72)y=00=-2165​(x−72)(x+72)LHS⋅(-5216​)=RHS⋅(-5216​)0=(x−72)(x+72) To solve this equation, we will apply the Zero Product Property. (x−72)(x+72)=0Use the Zero Product Propertyx−72=0x+72=0​(I)(II)​(I):  LHS+72=RHS+72x=72x+72=0​(II):  LHS−72=RHS−72x1​=72x2​=-72​ The width is the distance between the x-coordinates, 72 and -72. ∣72−(-72)∣=144​ It is 144 inches.