But the $$y$$-intercept is at $$(0,-2)$$, so we have to solve for $$a$$: $$\displaystyle -2=a\left( {0-3} \right){{\left( {0+1} \right)}^{2}};\,\,\,\,\,-2=a\left( {-3} \right)\left( 1 \right);\,\,\,\,\,\,\,a=\frac{{-2}}{{-3}}\,\,=\,\,\frac{2}{3}$$, The polynomial is $$\displaystyle y=\frac{2}{3}\left( {x-3} \right){{\left( {x+1} \right)}^{2}}$$. 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In fact, you can even put in, First use synthetic division to verify that, Subtract down, and bring the next digit (, $$x$$  goes into $$\displaystyle {{x}^{3}}$$ $$\color{red}{{{{x}^{2}}}}$$ times, Multiply the $$\color{red}{{{{x}^{2}}}}$$ by “$$x+3$$ ” to get $$\color{red}{{{{x}^{3}}+3{{x}^{2}}}}$$, and put it under the $${{x}^{3}}+7{{x}^{2}}$$. (Hint: Each side of the three-dimensional box has to have a length of at least, (c) Find the value of $$x$$ for which $$V\left( x \right)$$ has the greatest volume. Draw a sign chart with critical values –3, 0, and 3. We also have 2 changes of signs for $$P\left( {-x} \right)$$, so there might be 2 negative roots, or there might be 0 negative roots. When P(x) = 0, the company's profit is 0, and we found that this happens when x = 25, or when 25 products are made and sold. Since volume is $$\text{length }\times \text{ width }\times \text{ height}$$, we can just multiply the three terms together to get the volume of the box. For b < -2 the parabola will intersect the x-axis in two points with positive x values (i.e. We've already seen the property that these values of x are the values that make a function equal to zero. Furthermore, take a close look at the Venn diagram below showing the difference between a monomial and a polynomial. Multiply all the factors to get Standard Form: $$\displaystyle y={{x}^{3}}+12{{x}^{2}}-60x+64$$. We want the negative intervals, not including the critical values. Since the remaining term is not factorable, use the Quadratic Formula to find another root. $$V\left( x \right)=\left( {2x+5} \right)\left( {2x} \right)\left( {2x+3} \right)$$, \begin{align}V\left( x \right)&=\left( {2x+5} \right)\left( {2x} \right)\left( {2x+3} \right)\\&=\left( {2x+5} \right)\left( {4{{x}^{2}}+6x} \right)\\&=8{{x}^{3}}+12{{x}^{2}}+20{{x}^{2}}+30x\\V\left( x \right)&=8{{x}^{3}}+32{{x}^{2}}+30x\end{align}, \begin{align}V\left( x \right)&=\left( {x+1} \right)\left( {2x} \right)\left( {x+3} \right)\\&=\left( {x+1} \right)\left( {2{{x}^{2}}+6x} \right)\\V\left( x \right)&=2{{x}^{3}}+8{{x}^{2}}+6x\end{align}. So for example, for the factored polynomial $$y=2x{{\left( {x-4} \right)}^{2}}{{\left( {x+8} \right)}^{3}}$$, the factors are $$x$$ (root 0 with multiplicity 1), $$x-4$$ (root 4 with multiplicity 2), and $$x+8$$ (root –8 with multiplicity 3). The solution is $$[-4,-1]\cup \left[ {3,\,\infty } \right)$$. Also note that sometimes we have to factor the polynomial to get the roots and their multiplicity. $$\begin{array}{c}30-2x>0;\,\,\,\,\,\,x<15\\15-2x>0;\,\,\,\,\,\,\,x<7.5\\x>0\end{array}$$ Domain: $$\left( {0,\,7.5} \right)$$. : a) From above, volume of the box in Factored Form is: b) To get the volume of the box remaining, just subtract the two volumes: We need to subtract two polynomials to get the volume of the box without the cutout section. Since we have a factor of $$\left( {x-2} \right)$$, multiplicity, Since the coefficient of the divisor is not, \displaystyle \begin{align}\frac{{\pm 1,\,\,\,\pm 2}}{{\pm 1,\,\,\,\pm 3}}\,\,&=\,\,\frac{{\pm 1,\,\,\,\pm 2}}{{\pm 1}},\,\,\,\frac{{\pm 1,\,\,\,\pm 2}}{{\pm 3}}\\\\&=\,\,\pm \,\,1,\,\,\pm \,\,2,\,\,\pm \,\,\frac{1}{3},\,\,\pm \,\,\frac{2}{3}\end{align}, \require{cancel} \displaystyle \begin{align}\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 4,\,\,\pm 8}}{{\pm 1,\,\,\,\pm 2}}\,&=\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 4,\,\,\pm 8}}{{\pm 1}},\,\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 4,\,\,\pm 8}}{{\,\,\,\pm 2}}\,\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\,\,\pm \,\,1,\,\,\pm \,\,2,\,\,\pm \,\,4,\,\,\pm \,\,8,\,\,\pm \,\,\frac{1}{2},\,\,\cancel{{\pm \,\,1}},\cancel{{\pm \,\,2}},\cancel{{\pm \,\,4}}\end{align}, \displaystyle \begin{align}\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\pm 1,\,\,\,\pm 2,\,\,\pm 3,\,\,\,\pm 4,\,\,\pm 6,\,\,\,\pm 12}}\,\,&=\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\pm 1}},\,\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\,\,\,\pm 2}}\\&=\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\pm 3}},\,\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\,\,\,\pm 4}}\\&=\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\pm 6}},\,\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\,\,\,\pm 12}}\\\,\,&=\,\,\pm \,\,1,\,\,\pm \,\,2,\,\,\pm \,\,3,\,\,\pm \,\,6,\,\,\pm \,\,\frac{1}{2},\,\,\pm \,\,\frac{3}{2},\\\,\,\,\,\,\,\,&\pm \,\,\frac{1}{3},\,\,\pm \,\,\frac{2}{3},\,\,\pm \,\,\frac{1}{4},\,\,\pm \,\,\frac{3}{4},\,\,\pm \,\,\frac{1}{6}\,\,,\,\,\pm \,\,\frac{1}{{12}}\end{align}. Anytime we're asked to find the zeros, roots, or x-intercepts of a function f(x), we're being asked to find what values of x make f zero. Let’s first talk about the characteristics we see in polynomials, and then we’ll learn how to graph them. This equation is equivalent to. *Note that there’s another (easier) way to find a factored form for a polynomial, given an irrational root (and thus its conjugate). {\underline {\, You can also hit WINDOW and play around with the Xmin, Xmax, Ymin and Ymax values. Concave downward. Multiply $$\color{blue}{{4x}}$$ by “$$x+3$$ ” to get $$\color{blue}{{4{{x}^{2}}+12x}}$$, and put it under the $$\displaystyle 4{{x}^{2}}+10x$$. The solution of a polynomial equation, f(x), is the point whose root, r, is the value of x when f(x) = 0.Confusing semantics that are best clarified with a few simple examples. (See how we get the same zeros?) Using the example above: $$2+3i$$ is a root, so let $$x=2+3i$$ or $$x=2-3i$$ (both get same result). Okay, now that we know what zeros, roots, and x-intercepts are, let's talk about some of their many properties. The solution is $$[-4,-1]\cup \left[ {3,\,\infty } \right)$$. If $$x-c$$ is a factor, then $$c$$ is a root (more generally, if $$ax-b$$ is a factor, then $$\displaystyle \frac{b}{a}$$ is a root.). From counting through calculus, making math make sense! Here are some questions that you might see on Factor or Remainder Theorems: $$P\left( x \right)=2{{x}^{4}}+6{{x}^{3}}+5{{x}^{2}}-45$$. For graphing the polynomials, we can use what we know about end behavior. There is a relative (local) minimum at $$5$$, where $$x=0$$. Roots and zeros When we solve polynomial equations with degrees greater than zero, it may have one or more real roots or one or more imaginary roots. To cover cost, the company must sell at least 25 products. (Sometimes the graphing calculator will display a very small number, instead of an actual 0.) (This is the zero product property: if $$ab=0$$, then $$a=0$$ and/or $$b=0$$). Subtract down, and bring the next term ($$+10x$$) down. eval(ez_write_tag([[728,90],'shelovesmath_com-medrectangle-3','ezslot_2',109,'0','0']));Think of a polynomial graph of higher degrees (degree at least 3) as quadratic graphs, but with more twists and turns. However, it doesn’t make a lot of sense to use this test unless there are just a few to try, like in the first case above. The answer is $$\left[ {- 2,2} \right]$$. Not all functions have end behavior defined; for example, those that go back and forth with the $$y$$ values and never really go way up or way down (called “periodic functions”) don’t have end behaviors. Then I moved the cursor to the right of the top after “Right Bound?”, and hit enter twice to get the maximum point. There is an absolute maximum (highest of the whole graph) at about at $$8.34$$, where $$x=-1.20$$ and a relative (local) maximum at about $$6.23$$, where $$x=.83$$. eval(ez_write_tag([[728,90],'shelovesmath_com-mobile-leaderboard-1','ezslot_21',112,'0','0']));The DesCartes’ Rule of Signs will tell you the number of positive and negative real roots of a polynomial $$P\left( x \right)$$ by looking at the sign changes of the terms of that polynomial. If $$P\left( x \right)=2{{x}^{4}}+6{{x}^{3}}+k{{x}^{2}}-45$$. Find the other zeros for the following function, given $$5-i$$ is a root: Two roots of the polynomial are $$i$$ and. The polynomial is $$y=2\left( {x+\,\,3} \right){{\left( {x+1} \right)}^{2}}{{\left( {x-1} \right)}^{3}}$$. 14 MULTIPLE ROOTS POINT OF INFLECTION W HEN WE STUDIED quadratic equations, we saw what it means for a polynomial to have a double root.. When you do these, make sure you have your eraser handy! It costs the makeup company, (a) Write a function of the company’s profit $$P$$, by subtracting the total cost to make $$x$$, kits from the total revenue (in terms of $$x$$, End Behavior of Polynomials and Leading Coefficient Test, Putting it All Together: Finding all Factors and Roots of a Polynomial Function, Revisiting Factoring to Solve Polynomial Equations, $$t\left( {{{t}^{3}}+t} \right)={{t}^{4}}+{{t}^{2}}$$, $$\displaystyle \frac{{\left( {x+4} \right)}}{2}+\frac{{xy}}{{\sqrt{3}}}+3$$, $$4{{x}^{3}}{{y}^{4}}+2{{x}^{2}}y+xy+3xy+x+y-4$$, $$x{{\left( {x+4} \right)}^{2}}{{\left( {x-3} \right)}^{5}}$$. If we can factor polynomials, we want to set each factor with a variable in it to 0, and solve for the variable to get the roots. Then check each interval with a sample value in the last inequality above and see if we get a positive or negative value. Let’s just evaluate the polynomial for $$x=-3$$: To get the remainder, we can either evaluate $$P\left( 3 \right)$$ (which is easier! This is why we also call zeros of a function x-intercepts of a function. Let's take a look at a geometric property of these values. You start out at your house and travel an out-and-back route, ending where you started - at your house. Now let’s factor what we end up with: $${{x}^{3}}+4{{x}^{2}}+x+4={{x}^{2}}\left( {x+4} \right)+1\left( {x+4} \right)=\left( {{{x}^{2}}+1} \right)\left( {x+4} \right)$$. the original equation will have two real roots, both positive). Aha! Construct a table of at least 4 ordered pairs of points on the graph of the following equation and use the ordered pairs from the table to sketch the graph of the equation. You can test out of the We are only talking about real roots; imaginary roots have similar curve behavior, but don’t touch the $$x$$-axis. Let’s try –2 for the leftmost interval: $$\left( {-3-2} \right)\left( {-3+2} \right)\left( {{{{\left( {-3} \right)}}^{2}}+1} \right)=\left( {-5} \right)\left( {-1} \right)\left( {10} \right)=\text{ positive (}+\text{)}$$. Let's see how that works. The zeros are $$5-i,\,\,\,5+i$$ and 5. which is $$y=a\left( {x-4} \right)\left( {{{x}^{2}}-2x-2} \right)$$* (distribute and multiply through the last two factors). You start out at your house and travel an out and back route, ending where you started - at your house. Since we weren’t given a $$y$$-intercept, we can take the liberty to write the polynomial with integer coefficients: $$P(x)=x\left( {3x+10} \right)\left( {4x-3} \right)$$. What is the Difference Between Blended Learning & Distance Learning? j. The leading coefficient of the polynomial is the number before the variable that has the highest exponent (the highest degree). | 1 We worked with Linear Inequalities and Quadratic Inequalities earlier. (a) Write a function of the company’s profit $$P$$ by subtracting the total cost to make $$x$$ kits from the total revenue (in terms of $$x$$). We have 1 change of signs for $$P\left( x \right)$$, so there might be 1 positive root. The graph of the polynomial above intersects the x-axis at x=-1, and at x=2.Thus it has roots at x=-1 and at x=2. The root of the word "vocabulary," for example, is voc, a Latin root meaning "word" or "name." Pretty cool! We may even have to factor out any common factors and then do some “unfoiling” or other type of factoring (this has a difference of squares): $$y=-{{x}^{4}}+{{x}^{2}};\,\,\,\,\,y=-{{x}^{2}}\left( {{{x}^{2}}-1} \right);\,\,\,\,\,y=-{{x}^{2}}\left( {x-1} \right)\left( {x+1} \right)$$. (Note that there’s another (easier) way to find a factored form for a polynomial, given an irrational root, and thus its conjugate. We have to set the new volume to twice this amount, or 120 inches. When 25 products are sold, revenue = cost. (Always. $$\left( {3-2} \right)\left( {3+2} \right)\left( {{{{\left( 3 \right)}}^{2}}+1} \right)=\left( 1 \right)\left( 5 \right)\left( {10} \right)=\text{ positive (}+\text{)}$$. For example, if you have the polynomial $$f\left( x \right)=-{{x}^{4}}+5{{x}^{3}}+2{{x}^{2}}-8$$, and if you put a number like 3 in for $$x$$, the value for $$f(x)$$ or $$y$$ will be the same as the remainder of dividing $$-{{x}^{4}}+5{{x}^{3}}+2{{x}^{2}}-8$$ by $$(x-3)$$. Do the same to get the minimum, but use 2nd TRACE (CALC), 3 (minimum). Illustrated definition of Polynomial: A polynomial can have constants (like 4), variables (like x or y) and exponents (like the 2 in ysup2sup),... A polynomial can have constants (like 4), variables (like x or y) and exponents (like the 2 in y 2), that can be combined using addition, subtraction, multiplication and division, but: The polynomial will thus have linear factors (x+1), and (x-2).Be careful: This does not determine the polynomial! We learned what a Polynomial is here in the Introduction to Multiplying Polynomials section. Now we need to find a different root for the equation $$P\left( x \right)=-4{{x}^{3}}+25x-24$$. Factors are $$3,x,\left( {x-2} \right),\text{and}\left( {{{x}^{2}}+2x+4} \right)$$, and real roots are $$0$$ and $$2$$ (we don’t need to worry about the $$3$$, and $${{x}^{2}}+2x+4$$ doesn’t have real roots). Remember that, generally, if $$ax-b$$ is a factor, then $$\displaystyle \frac{b}{a}$$ is a root. When we want to factor and get the roots of a higher degree polynomial using synthetic division, it can be difficult to know where to start! {\,0\,\,\,} \,}} \right. {\,\,1\,\,} \,}}\! Since $$1-\sqrt{3}$$ is a root, by the conjugate pair theorem, so is $$1+\sqrt{3}$$. The polynomial is $$\displaystyle y=\frac{1}{4}\left( {x-4} \right)\left( {{{x}^{2}}-2x-2} \right)$$. Round to 2 decimal places. To do this we set the polynomial to zero in the form of an equation: Then we just solve the equation. The factor that represents these roots is $${{x}^{2}}-4x+13$$. Definition of root as used in math 1. Notice that we have 3 real solutions, two of which pass through the $$x$$-axis, and one “touches” it or “bounces” off of it: Notice also that each factor has an odd exponent when the graph passes through the $$x$$-axis and an even exponent when the function “bounces” off of the $$x$$-axis. We have to be careful to either include or not include the points on the $$x$$-axis, depending on whether or not we have inclusive ($$\le$$ or $$\ge$$) or non-inclusive ($$<$$ and $$>$$) inequalities. flashcard set, {{courseNav.course.topics.length}} chapters | $$f\left( x \right)={{x}^{4}}+{{x}^{3}}-3{{x}^{2}}-x+2$$, $$\displaystyle \pm \frac{p}{q}\,=\,\pm \,1,\,\,\pm \,2$$. (b) What would be a reasonable domain for the polynomial? Note that in the second example, we say that $${{x}^{2}}+4$$ is an irreducible quadratic factor, since it can’t be factored any further (therefore has imaginary roots). From earlier we saw that “–3” is a root; this is the negative root. The company could sell 1.386 thousand or 1,386 kits and still make the same profit as when it makes 1500 kits. We learned that a Quadratic Function is a special type of polynomial with degree 2; these have either a cup-up or cup-down shape, depending on whether the leading term (one with the biggest exponent) is positive or negative, respectively. Remember that the degree of the polynomial is the highest exponent of one of the terms (add exponents if there are more than one variable in that term). For example, the end behavior for a line with a positive slope is: $$\begin{array}{l}x\to -\infty \text{, }\,y\to -\infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$$, and the end behavior for a line with a negative slope is: $$\begin{array}{l}x\to -\infty \text{, }\,y\to \infty \\x\to \infty \text{, }\,\,\,y\to -\infty \end{array}$$. \right| \,\,\,\,\,1\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,-15\,\,\,\,\,\,-10\,\,\,\,\,\,\,\,\,\,\,\,\,\,k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,72\\\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9\,\,\,\,\,\,\,-18\,\,\,\,\,\,\,\,-84\,\,\,\,\,\,\,\,\,\,3\left( {k-84} \right)\,\,\,\,\,\,\,\,\,\,}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,-6\,\,\,\,\,\,\,-28\,\,\,\,\,\,\,k-84\,\,\,\,\left| \! Round to, that a makeup company can charge for a certain kit is $$p=40-4{{x}^{2}}$$, where $$x$$ is the number (in thousands) of kits produced. Since the $$y$$-intercept is at $$(0,2)$$, let’s solve for $$a$$: $$\displaystyle 2=a\left( {0-4} \right)\left( {{{0}^{2}}-2\left( 0 \right)-2} \right);\,\,\,\,2=8a;\,\,\,\,a=\frac{1}{4}$$. Use open circles for the critical values since we have a $$<$$ and not a $$\le$$ sign. What Are Roots? 0 Comments Show Hide all comments Sign in to comment. Multiply all the factors to get Standard Form: $$y=2{{x}^{4}}-16{{x}^{3}}+46{{x}^{2}}-64x-130$$. Note that the value of $$x$$ at the highest point is, We can put the polynomial in the graphing calculator using either the standard or factored form. {{courseNav.course.mDynamicIntFields.lessonCount}} lessons You might have to go backwards and write an equation of a polynomial, given certain information about it: $$\begin{array}{c}x\to -\infty \text{, }\,y\to -\infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$$. Graph of a cubic function with 3 real roots (where the curve crosses the horizontal axis—where y = 0).The case shown has two critical points.Here the function is f(x) = (x 3 + 3x 2 − 6x − 8)/4. The rational root test help us find initial roots to test with synthetic division, or even by evaluating the polynomial to see if we get 0. Round to 2 decimal places. The square root function defined above is evaluated for some nonnegative values of xin the table below. Multiply all the factors to get Standard Form: $$\displaystyle y=\frac{1}{4}{{x}^{3}}-\frac{3}{2}{{x}^{2}}+\frac{3}{2}x+2$$. graph /græf/ USA pronunciation n. []a diagram representing a system of connections or relations among two or more things, as by a number of Factors are $$\left( {3x+2} \right),\left( {x-5} \right),\,\text{and}\left( {x+1} \right)$$, and roots are $$\displaystyle -\frac{2}{3},5,\text{and}-1$$. $$\boldsymbol{y}$$-intercept: Note that the $$y$$-intercept of the polynomial function (when $$x=0$$) is $$(0,–12)$$. No coincidence here either with its end behavior, as we’ll see. These values have a couple of special properties. Let's consider another example of how zeros, roots, and x-intercepts can give us a whole bunch of information about a function. Visit the Honors Precalculus Textbook page to learn more. Anyone can earn Compare the nature of roots to the actual roots: Here is a graph of the above equation. (negative coefficient, even degree), we can see that the polynomial should have an end behavior of $$\begin{array}{l}x\to -\infty \text{, }\,y\to -\infty \\x\to \infty \text{, }\,\,\,y\to -\infty \end{array}$$, which it does! Interactive lesson on the graph of y = a(x − h)² + k, for positive and negative a, including the number of roots, using sliders. To learn more, visit our Earning Credit Page. {\overline {\, We can solve these inequalities either graphically or algebraically. (a) Profit is total revenue to make all $$x$$ thousand kits minus the cost to make all $$x$$ thousand kits. Go down a level (subtract 1) with the exponents for the variables: $$\begin{array}{l}\color{blue}{1}{{x}^{2}}+\color{brown}{4}x\color{purple}{{-2}}\\\,={{x}^{2}}+4x-2\end{array}$$. The solution is $$\left( {-3,0} \right)\cup \left( {0,3} \right)$$, since we can’t include 0, because of the $$<$$. Let’s try some problems, and solve both graphically and algebraically: $$-\left( {x+1} \right)\left( {x+4} \right)\left( {x-3} \right)\le 0$$. • Below is the graph of a polynomial q(x). Here are examples (assuming we can’t use a graphing calculator to check for roots). In these examples, one of the factors or roots is given, so the remainder in synthetic division should be 0. But sometimes "root" is used as a quick way of saying "square root", for example "root 2" means √2. It says: $$P\left( x \right)={{x}^{4}}+{{x}^{3}}-3{{x}^{2}}-x+2$$, $$P\left( x \right)\,\,=\,\,+\,{{x}^{4}}\color{red}{+}{{x}^{3}}\color{red}{-}3{{x}^{2}}\color{lime}{-}x\color{lime}{+}2$$. Now, perform the synthetic division, using the fractional root (see left)! We want $$\le$$ from the factored inequality, so we look for the – (negative) sign intervals, so the interval is $$\left[ {- 2,2} \right]$$. $$10{{x}^{4}}-13{{x}^{3}}-21{{x}^{2}}+10x+8$$. So, to get the roots (zeros) of a polynomial, we factor it and set the factors to 0. The roots (zeros) are $$-1+\sqrt{7},\,\,-1-\sqrt{7},\,\,-3$$, and $$1$$. writing Examples of words with the root -graph: lithograph Abused, Confused, & Misused Words by … Notice that we can use synthetic division again by guessing another factor, as we do in the last problem: Factors are $$\left( {x+3} \right),\left( {5x+6} \right),\text{and}\left( {x-3} \right)$$, and real roots are $$\displaystyle -3,-\frac{6}{5},\text{and}\,3$$. So when you want to find the roots of a function you Let’s multiply out to get Standard Form and set to 120 (twice the original volume). $$P\left( x \right)={{x}^{5}}-15{{x}^{3}}-10{{x}^{2}}+kx+72$$. Those are roots and x-intercepts. (We’ll learn about this soon). Log in or sign up to add this lesson to a Custom Course. We used vertical multiplication for the polynomials: $$\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}+9x+20\\\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\times \,\,\,\,\,x\,\,+3}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,3{{x}^{2}}+27x+60\\\underline{{{{x}^{3}}+\,\,\,9{{x}^{2}}+20x\,\,\,\,\,\,\,\,\,\,\,\,\,}}\\{{x}^{3}}+12{{x}^{2}}+47x+60\end{array}$$. Graph Theory: Penn State Math 485 Lecture Notes Version 1.5 Christopher Gri n « 2011-2020 Licensed under aCreative Commons Attribution-Noncommercial-Share Alike 3.0 United States License With Contributions By: Elena The Factor Theorem basically repeats something that we already know from above: if a number is a root of a polynomial (like if 3 is a root of $${{x}^{2}}-9$$, which it is), then when you divide 3 into $${{x}^{2}}-9$$ (like with synthetic division), you get a remainder of 0. -4\ ) minimum ) is 1 ( which is actually its exponent )... 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