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Exercises 1 We want to find a difference, identify the degree of this difference and classify it by the number of terms.Finding a Difference The first step in finding a difference is to 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. (-2p+4)−(p2−6p+8)​ In this case, we have one p2-term, two p-terms, and two constants. Only the p-terms and constants can be combined, so to simplify the expression we will use the Distributive Property and rearrange it according to the Commutative Property of Addition and then combine like terms. (-2p+4)−(p2−6p+8)Distribute -1(-2p+4)−p2+6p−8Remove parentheses-2p+4−p2+6p−8Commutative Property of Addition-p2−2p+6p+4−8Add and subtract terms-p2+4p−4Identifying the Degree The degree of a polynomial is the highest exponent of a variable. -p2+4p−4​ The highest exponent of the polynomial, and therefore its degree, is 2.Classifying We want to classify this polynomial by the number of terms. -p2+4p−4​ This polynomial consists of three terms, so it is a trinomial.
Exercises 2 We want to find a difference, identify the degree of this difference and classify it by the number of terms.Finding a Difference The first step in finding a difference is to 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. (9c6−5b4)−(4c6−5b4)​ In this case, we have two c6-terms, and two b4-terms. All terms can be combined, so to simplify the expression we will use the Distributive Property and rearrange it according to the Commutative Property of Addition and then combine like terms. (9c6−5b4)−(4c6−5b4)Distribute -1(9c6−5b4)−4c6+5b4Remove parentheses9c6−5b4−4c6+5b4Commutative Property of Addition9c6−4c6−5b4+5b4Add and subtract terms5c6Identifying the Degree The degree of a polynomial is the highest exponent of the variable. 5c6​ The highest exponent of the polynomial, and therefore its degree, is 6.Classifying We want to classify this polynomial by the number of terms. 5c6​ This polynomial consists of one term, so it is a monomial.
Exercises 3 We want to find a sum, identify the degree of this addition and classify it by the number of terms.Finding a Sum The first step in finding a sum is to 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. (4s4+2st+t)+(2s4−2st−4t)​ In this case, we have two s4-terms, two st-terms, and two t-terms. All terms can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then combine like terms. (4s4+2st+t)+(2s4−2st−4t)Remove parentheses4s4+2st+t+2s4−2st−4tCommutative Property of Addition4s4+2s4+2st−2st+t−4tAdd and subtract terms6s4−3tIdentifying the Degree The degree of a polynomial is the highest exponent of a variable. 6s4−3t​ The highest exponent of the polynomial, and therefore its degree, is 4.Classifying We want to classify this polynomial by the number of terms. 6s4−3t​ This polynomial consists of two terms, so it is a binomial.
Exercises 4 We want to find a product. To do so, we will apply the Distributive Property. (h−5)(h−8)Distribute (h−8)h(h−8)−5(h−8)Distribute hh2−8h−5(h−8)Distribute -5h2−8h−5h+40Subtract termh2−13h+40
Exercises 5 We want to find a product. To do so, we will apply the Distributive Property. (2w−3)(3w+5)Distribute (3w+5)2w(3w+5)−3(3w+5)Distribute 2w6w2+10w−3(3w+5)Distribute -36w2+10w−9w−15Subtract term6w2+w−15
Exercises 6 We are ask the expand the product of binomials (z+11)(z−11). To do this, notice that it is the product of the sum and difference of two terms. Therefore, we can can know that it will result in the difference of two squares and we can use the corresponding pattern. (a+b)(a−b)=a2−b2​ Let's give it a try. (z+11)(z−11)(a+b)(a−b)=a2−b2z2+112Calculate powerz2+121
Exercises 7 We can start by reviewing what a perfect square trinomial is and how can we obtain one. Recall that the process of squaring a binomial gives as a result a perfect square trinomial. Let's take a look at the corresponding pattern. (a+b)2=a2+2ab+b2​(a−b)2=a2−2ab+b2​​ From the patterns we can see that a trinomial has to satisfy two conditions to be a perfect square trinomial.Two of its terms have to be perfect squares. This is represented in the pattern by the terms a2 and b2. The other term has to equal two times the product of the square roots of the other terms. This is represented in the pattern by the term 2ab.To illustrate this, let's consider the example trinomial shown below. x2+8x+16​ Note that x2 and 16=42 are perfect squares whose square roots are x and 4. Hence, the first condition is satisfied. Furthermore, the middle term 8x equals two times the product of the square roots of the other terms. 8x 8x 8x ​= 2⋅x⋅4= 2⋅4⋅x= 8x✓​ The second condition is also satisfied and we can know that the trinomial x2+8x+16 is a perfect square trinomial.
Exercises 8 Let's start by reviewing the definition of a polynomial.A polynomial is a monomial or sum of monomials, where each of the monomials comprising it is called a term of the polynomial. As we can see, the definition of polynomial does not restrict the amount of terms it can have. Furthermore, as it is comprised of monomials and it is possible for a monomial to be a constant term, 18 is an example of a constant polynomial.
Exercises 9 To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term. s2−15s+50​ In this case, we have 50. This is a positive number, so for the product of the constant terms in the factors to be positive, these constants must have the same sign (both positive or both negative.)Factor ConstantsProduct of Constants 1 and 5050 -1 and -5050 2 and 2550 -2 and -2550 5 and 1050 -5 and -1050 Next, let's consider the coefficient of the linear term. s2−15s+50​ For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, -15.FactorsSum of Factors 1 and 5051 -1 and -50-51 2 and 2527 -2 and -25-27 5 and 1015 -5 and -10-15We found the factors whose product is 50 and whose sum is -15. s2−15s+50⇔(s−5)(s−10)​ Checking Our Answer info Check your answer ✓ We can check our answer by applying the Distributive Property and comparing the result with the given expression. (s−5)(s−10)Distribute (s−10)s(s−10)−5(s−10)Distribute ss2−10s−5(s−10)Distribute -5s2−10s−5s+50Subtract terms2−15s+50✓ After applying the Distributive Property and simplifying, the result is the same as the given expression. Therefore, we can be sure our solution is correct!
Exercises 10 To factor by grouping, we will take the greatest common factor of the first pair of terms and the greatest common factor of the second pair of terms. h3+2h2−9h−18Factor out h2h2(h+2)−9h−18Factor out -9h2(h+2)−9(h+2)Factor out (h+2)(h+2)(h2−9) To factor h2−9, we will apply the difference of squares rule. (h+2)(h2−9)Write as a power(h+2)(h2−32)a2−b2=(a+b)(a−b)(h+2)(h+3)(h−3)
Exercises 11 Here we have a quadratic trinomial of the form ax2+bx+c, where ∣a∣​=1 and there are no common factors. To factor this expression, we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. -5k2−22k+15⇔-5k2+(-22)k+15​ We have that a=-5, b=-22, and c=15. There are now three steps we need to follow in order to rewrite the above expression.Find ac. Since we have that a=-5 and c=15, the value of ac is -5×15=-75. Find factors of ac. Since ac=-75, which is negative, we need factors of ac to have opposite signs — one positive and one negative — in order for the product to be negative. Since b=-22, which is also negative, the absolute value of the negative factor will need to be greater than the absolute value of the positive factor, so that their sum is negative.1st factor​135​2nd factor​-75-25-15​Sum​1+(-75)3+(-25)5+(-15)​Result​-74-22-10​Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. -5k2+(-22)k+15⇕-5k2+3k−25k+15​Finally, we will factor the last expression obtained. -5k2+3k−25k+15Factor out kk(-5k+3)−25k+15Factor out 5k(-5k+3)+5(-5k+3)Factor out (-5k+3)(-5k+3)(k+5)Commutative Property of Addition(3−5k)(k+5) Checking Our Answer info Check your answer ✓ We can expand our answer and compare it with the given expression. (3−5k)(k+5)Distribute (k+5)3(k+5)−5k(k+5)Distribute 33k+15−5k(k+5)Distribute -5k3k+15−5k2−25kCommutative Property of Addition-5k2+3k−25k+15Subtract term-5k2−22k+15✓ We can see above that after expanding and simplifying, the result is the same as the given expression. Therefore, we can be sure our solution is correct!
Exercises 12 Since the function is already written in factored form, we will use the Zero Product Property. (n−1)(n+6)(n+5)=0Use the Zero Product Propertyn−1=0n+6=0n+5=0​(I)(II)(III)​(I):  LHS+1=RHS+1n=1n+6=0n+5=0​(II):  LHS−6=RHS−6n=1n=-6n+5=0​(III):  LHS−5=RHS−5n=1n=-6n=-5​ We found that the zeros of the function are 1, -6, and -5.
Exercises 13 To solve the given equation by factoring, we will have to factor a perfect square trinomial. a2+2ab+b2⇔(a+b)2​Factoring the Expression For the given equation, all the terms are on the left-hand side and there is no greatest common factor (GCF) to factor out first, so let's start factoring. d2+14d+49=0Split into factorsd2+2(7)d+49=0Commutative Property of Multiplicationd2+2d(7)+49=0Write as a powerd2+2d(7)+72=0a2+2ab+b2=(a+b)2(d+7)2=0Solving the Equation Now let's solve the equation by taking square roots. (d+7)2=0LHS​=RHS​d+7=0LHS−7=RHS−7d=-7 The solution to this equation is d=-7.
Exercises 14 We want to solve the given equation by factoring.Factoring Let's start by writing all the terms on one side of the equals sign. We'll also factor out a greatest common factor (GCF), if we find one. 6x4+8x2=26x3LHS−26x3=RHS−26x36x4+8x2−26x3=0Factor out 2x22x2(3x2+4−13x)=0Commutative Property of Addition2x2(3x2−13x+4)=0Write as a difference2x2(3x2−12x−x+4)=0 Factor out 3x & -1 Factor out 3x2x2(3x(x−4)−x+4)=0Factor out -1 2x2(3x(x−4)−1(x−4))=0Factor out (x−4)2x2(x−4)(3x−1)=0Solving To solve this equation, we will apply the Zero Product Property. 2x2(x−4)(3x−1)=0Use the Zero Product Property2x2=0x−4=03x−1=0​(I)(II)(III)​(I):  LHS/2=RHS/2x2=0x−4=03x−1=0​(I):  LHS​=RHS​x=0x−4=03x−1=0​(II):  LHS+4=RHS+4x=0x=43x−1=0​(III):  LHS+1=RHS+1x=0x=43x=1​(III):  LHS/3=RHS/3x1​=0x2​=4x3​=31​​
Exercises 15 We are asked to write a polynomial in standard form. We will get it by factoring the given polynomial completely and distributing π. π(r−3)2(a−b)2=a2−2ab+b2π(r2−2(r)(3)+32)Multiplyπ(r2−6r+32)Calculate powerπ(r2−6r+9)Distribute ππr2−6πr+9π
Exercises 16
Exercises 17 We are asked to find a polynomial equation with three positive roots. If in the factored form a polynomial has a factor in a form (x−r), it means that r is the root of the polynomial. So, a polynomial equation with three roots r1​, r2​, and r3​, can be in the following form. (x−r1​)(x−r2​)(x−r3​)=0​ Let's choose r1​=1, r2​=2, and r3​=3 to be the positive roots of a polynomial equation. (x−r1​)(x−r2​)(x−r3​)=0⇓(x−1)(x−2)(x−3)=0​ Notice that instead of 1, 2, and 3, we could choose any three other positive numbers to be the roots of a polynomial equation.
Exercises 18 Let's plot the graph of the given function, y=-16t2+24t. If you need explanations on graphing the quadratic function, please refer to here.At the time t=0 we are at level y=0. Since the next time we will be at trampoline-level is when y=0, let's solve the equation y=0 for t. y=0y=-16t2+24t-16t2+24t=0 Solve for t Factor out tt(-16t+24)=0Use the Zero Product Propertyt=0-16t+24=0​(I)(II)​(II): LHS−24=RHS−24t=0-16t=-24​(I)(II)​(II): LHS/-16=RHS/-16t=0t=-16-24​​(I)(II)​(II): ba​=b/-8a/-8​ t=0t=23​​ Therefore, we are at trampoline-level for t=0 and t=23​. We are in the air exactly 23​, or 1.5, seconds.
Exercises 19