Zero product property
WELCOME TO A LESSON ON THE ZERO PRODUCT PROPERTY.
USE THE ZERO PRODUCT PROPERTY TO SOLVE POLYNOMIAL EQUATIONS IN FACTORING FORM. SO THE ZERO PRODUCT PROPERTY IS ONE OF THE MAIN REASONS WHY WE LEARN HOW TO FACTOR QUADRATIC EQUATIONS AND POLYNOMIAL EQUATIONS.
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THE ZERO PRODUCT PROPERTY STATES, “IF TWO NUMBERS, A AND B ARE MULTIPLIED TOGETHER “AND THE RESULTING PRODUCT IS ZERO, THEN AT LEAST ONE OF THE NUMBERS MUST BE ZERO.”
SO IF A x B = 0
THEN EITHER A MUST = 0 OR B MUST = 0 OR BOTH A AND B ARE = TO 0.
AND WE CAN USE THIS IDEA TO HELP US SOLVE POLYNOMIAL EQUATIONS IN FACTORED FORM. IF WE WANT TO SOLVE THE EQUATION X x THE QUANTITY X -3 = 0, BECAUSE THIS PRODUCT IS = TO 0 EITHER THE FIRST FACTOR OF X MUST = 0 OR THE SECOND FACTOR OF X -3 MUST = 0. SO WE KNOW ONE SOLUTION IS X = 0. AND THE SECOND SOLUTION WE HAVE TO SOLVE THIS EQUATION FOR X SO WE’D ADD 3 TO BOTH SIDES OF THE EQUATION. SO – 3 + 3 IS = TO 0. SO OUR SECOND SOLUTION IS X = +3. AGAIN, WE HAVE TWO SOLUTIONS. X = 0 OR X = 3.
OF COURSE, IF WE WANTED TO WE COULD CHECK THIS. TO CHECK X = 0 WE WOULD SUBSTITUTE 0 FOR X, WE WOULD HAVE 0 x THE QUANTITY 0 – 3. WELL, THAT WOULD BE 0 x -3 WHICH = 0. THAT CHECKS. AND WHEN X IS = TO 3 WE’D SUBSTITUTE 3 FOR X, WE WOULD HAVE 3 x 3 – 3. WELL, 3 – 3 IS = TO 0. SO HERE WE WOULD HAVE 3 x 0, WHICH ALSO = 0 AND THEREFORE CHECKS. LET’S TAKE A LOOK AT SOME MORE EXAMPLES.
zero product property calculator
IN THIS EQUATION WE HAVE 4 X x THE QUANTITY X + 5 = 0. AGAIN, BECAUSE THIS PRODUCT IS = TO 0 EITHER THE FIRST FACTOR OF 4 X MUST = 0 OR THE SECOND FACTOR OF X + 5 MUST = 0. AND NOW WE NEED TO SOLVE EACH OF THESE EQUATIONS FOR X TO DETERMINE OUR SOLUTIONS. SO HERE TO ISOLATE X WE WOULD DIVIDE BOTH SIDES BY 4. SO THIS WOULD BE 1 X OR JUST X = 0 DIVIDED BY 4 IS 0. TO SOLVE THIS EQUATION FOR X WE WOULD SUBTRACT 5 ON BOTH SIDES +5 – 5 IS = TO 0. SO WE HAVE X = -5. AGAIN, WE HAVE 2 SOLUTIONS, X = 0 OR X = -5. HERE WE HAVE THE QUANTITY X – 2 x THE QUANTITY X + 7 = 0. AGAIN, BECAUSE THIS PRODUCT IS = TO 0 EITHER X – 2 MUST = 0 OR X + 7 MUST = 0. AND NOW WE’LL SOLVE THESE EQUATIONS FOR X. SO HERE WE ADD 2 TO BOTH SIDES OF THE EQUATION, – 2 + 2 IS 0.
SO WE’RE LEFT WITH X = +2 OR SOLVING THIS EQUATION FOR X WE WOULD SUBTRACT 7 ON BOTH SIDES OF THE EQUATION WHICH WOULD GIVE US X = -7. SO THESE ARE THE 2 SOLUTIONS TO OUR POLYNOMIAL EQUATION IN FACTORED FORM. SO HOPEFULLY NOW YOU’RE BEGINNING TO SEE WHY IT’S BENEFICIAL TO HAVE A POLYNOMIAL EQUATION IN FACTORED FORM. LET’S TAKE A LOOK AT TWO MORE EXAMPLES. HERE WE HAVE THE QUANTITY 2 X + 3 x THE QUANTITY 5 X – 1 = 0. SO EITHER THE FIRST FACTOR OF 2 X + 3 MUST = 0 OR THE SECOND FACTOR OF 5 X – 1 MUST = 0. AND NOW WE’LL SOLVE THESE EQUATIONS FOR X. SO HERE WE WOULD START BY SUBTRACTING 3 ON BOTH SIDES OF THE EQUATION. THIS WOULD GIVE US 2 X = -3 AND THE LAST STEP HERE IS TO DIVIDE BOTH SIDES OF THE EQUATION BY 2. THIS WOULD BE 1 X OR X = -3 HALVES OR SOLVING THIS EQUATION FOR X WE WOULD START BY ADDING 1 TO BOTH SIDES OF THE EQUATION, THIS WOULD BE 0. SO IF 5 X EQUALS 1 AND DIVIDE BOTH SIDES BY 5.
SO OUR SECOND SOLUTION IS X = 1/5th. LET’S TAKE A LOOK AT ONE MORE EXAMPLE. NOTICE IN THIS EQUATION WE HAVE 3 FACTORS THAT HAVE A PRODUCT OF ZERO. SO IN THIS CASE WE’LL HAVE 3 SOLUTIONS EITHER X = 0FROM THIS FIRST FACTOR OR X – 1 = 0 FROM THE SECOND FACTOR OR 6 X + 11 IS = TO 0. AND NOW WE’LL SOLVE THESE FOR X. WELL, THE FIRST EQUATION IS ALREADY SOLVED FOR X. WE HAVE X = 0. THE SECOND EQUATION, WE’LL ADD 1 TO BOTH SIDES. THIS WILL GIVE US THE SOLUTION X = +1. AND THEN FOR THE THIRD EQUATION WE HAVE A 2 STEP EQUATION SO WE’LL SUBTRACT 11 ON BOTH SIDES AND THEN DIVIDE BOTH SIDES BY 6. SO OUR THIRD SOLUTION IS X = -11/6. SO AS LONG AS WE HAVE OUR PRODUCT = TO 0 WE CAN TAKE ADVANTAGE OF THE ZERO PRODUCT PROPERTY TO SOLVE THE POLYNOMIAL EQUATION. SO BECAUSE OF THE ZERO PRODUCT PROPERTY WE WILL SPEND SOME TIME LEARNING HOW TO FACTOR A VARIETY OF POLYNOMIALS SO THAT WE CAN SOLVE POLYNOMIAL EQUATIONS. I HOPE YOU FOUND THIS HELPFUL.
What the zero product rule
we will look at how we can solveequations with x squared by introducingthe zero product property. What the zero product rule says is that if you’re multiplyingtwo factors and the answer is 0, the only waythat’s possible is if either the first factor aequals 0, or the second factor b equals 0.
So if we have factorsthat equal 0, to solve,we set each factor… equal to 0. Let’s take a lookat some examples where we can set factors equalto 0 to find out what x equals. In this problem,we have two factors– 5x minus 1 and 2x plus 5–that multiply to equal 0. This means oneof the factors must be 0, so we write 5x minus 1 equals 0,or the 2x plus 5 equals 0, and solve these two equationsto find our solutions for x. The first equation we solveby adding 1: 5x equals 1; and finally dividing by 5. x could be equalto the fraction 1/5.
When it does,the first factor equals 0 times anythingis always 0. Similarly, we can solvethe second factor by subtracting 5from both sides. This gives us2x equals -5. Finally, dividingboth sides by 2, we find out xcould be the fraction -5/2. This means when x is -5/2,the second factor equals 0, times the first,will always be 0. Let’s take a lookat another example where we use this zero productproperty to find solutions. Here, we have three factorsmultiplied together: 2x is multiplied by x minus 6,is multiplied by 2x plus 3. One of those factors must be 0for this equation to equal 0, so we set them each equal to 0:2x equals 0, x minus 6 equals 0, and 2x plus 3 equals 0. Solving these equations will tell usthe possible values for x.
The first equation solvesby dividing by 2. This gives usx equals 0. When x equals 0, the first factoris 0 times something, times something,will always equal 0. We can solve the second equationby adding 6 to both sides. This gives usx equals 6. When x is equal to 6,the middle factor is 0, multiplied by the others,will always equal 0. To solve the third factor,we start by subtracting 3 to give us2x equals -3. Finally, dividing by 2gives us our final possible solutionfor x of -3/2. When x is -3/2,the last factor is 0, multiplied by the others,would always be 0. The zero product rulesimply says if several factorsmultiply to 0, either one of ’emor another will equal 0. Using this, we can findpossible solutions for x.
For the rest of the examples in thislesson, we’re just going to practice factoring together. If you struggle atall with factoring, I recommend you to download the steps for factoringtemplate that I have uploaded to Blackboard course. It will make thingseasier for you. So 9 x squared minus 81 equal to 0. There are two terms what Ialways start with no matter what is to find a GCF. In this case the largestnumber I could take out of both terms would be 9. When I do that I’m goingto divide each by 9. But I can’t just get rid of the 9. I can move it to the outside. Here thatwould leave me with x squared minus 9 equals to zero.
Then I would look for how to factorsomething that has two terms and in two terms I’m either finding a GCF or I’mfinding the difference of squares pattern and after that there’s reallynothing else. So I will in this case this does meet the difference of squarespattern x plus 3, x minus 3 and of course using that zero product property we said eachof these equal to 0 to find our solutions of negative 3 and positive3 and notice I’ve written that instead notation. If you can do the mathin your hand feel free to do the math in your head on the finding the zeros. Forthe second one this also has two terms but I wanted to show you the differencebetween the two. This has no middle value. So it was like plus 0 x. So, the bvalue is 0. In this case there’s no c value.
So it’s like plus zero at the endwhich is different. In this case when I’m finding my GCF it’s not just a 5 thatwill come out of both terms but a 5 x. So I moved the GCF to the outsideit’s 5 x which leaves me with x minus 5 left over. Now in this case,over here when I said each of these equal to 0 which I just did in my head, Isubtracted the 3 to get negative 3 I added the 3 to get 0. Here I do have to set the 5 x equal to 0. I didn’t have to do this here because of 9 did notcontain a variable 5 x does contain a variable. Therefore Ihave to set that one equal to 0. So that’s the difference between the two. Dividing by 5 ofcourse x is just zero. Adding 5, x is 5. So my two solutions are 0 and 5.
introduced quadratic functions and quadratic equations
So far we have introduced quadratic functions and quadratic equations. We have seen that the graph of aquadratic function of x is a parabola and by setting that function equal to zeroand solving the resulting quadratic equation we can find the values of xwhere the parabola intersects the x-axis in other words, the parabola’s “x-intercepts”. These values graphically correspondto the “zeros” or “roots” of the function.
So let’s take a look at the different waysthat a quadratic function of x can intersect the x-axis. We’ll start with the simplestquadratic function of x, “x-squared”. Setting this function equal to zero gives us the quadratic equation for this function. The only value of x which when squaredis equal to zero, is zero. So the quadratic function “x-squared”has a single x-intercept at the point where x equals zero. Now let’s do the same thing with a slightlymore complicated quadratic function “x-squared – 4”. Setting this function equal to zero gives us the quadratic equation”x-squared – 4″ equals zero or “x-squared equals 4”. There are two values of xwhich when squared are equal to 4 “positive 2” and “negative 2”. These two solutions can be written as”x equals plus or minus 2″ which corresponds to the two points wherethe function’s graph intersects the x-axis. So when the graph of a quadratic function of xintersects the x-axis it can have either one or two x-intercepts and thus one or two roots. It is also possible for a quadratic functionof a real variable to have no roots.
This situation occurs when its graphdoes not intersect the x-axis. However, this is true whenonly real numbers are considered. Several hundred years ago a new type of number called an “imaginarynumber” was created by mathematicians specifically because of functionswhich have no real roots. When imaginary numbers and “complex numbers”which include imaginary and real numbers are included in our number system it can be said that all quadratic functionshave two roots although these roots may or may not be real and may or may not be distinct. Later in these lectures, we will introduceimaginary and complex numbers and see how these numbers opened upa whole new world of mathematics. As we have just seen solving a quadratic equation of the form”x-squared minus c equals zero” is easy since as long as “c” has a positive value this simply amounts to findingthe square root of c. The same is true for equations of the form”a x-squared minus c equals zero “
since if we divide the equation by a we get the same type of equation as before except that we takethe square root of “c over a” instead of the square root of c. On the other hand, quadratic equationsthat include an x term can be more difficult to solve. There are a number of methodsfor solving quadratic equations like this. The methods we will examinein the next few lectures are based on the “zero product property”. So what exactly is the zero product property? Let’s say that we have a number or expression which is the product of two or more factors. For example, let’s assume that the productis composed of two factors which we will call A and B.
zero product property definition
The zero product property statesthat if any factor is equal to zero then the product of those factorsmust also be zero. This same logic applies to the linear factorsof a quadratic expression. As we showed in the lecture “Creating QuadraticExpressions Using the FOIL Method” a quadratic expression can be created bymultiplying two linear expressions together. Those linear expressions will therefore befactors of the resulting quadratic expression. So when x has a value that causes any factorof a quadratic expression to be zero then the quadratic expressionwill also have a value of zero. This seems obvious since the product ofanything multiplied by zero must be zero.
But the zero property also statesthat the converse is true. If a product of factors is zero then one or more of the factors must be zero. So every zero of a quadratic function must also be a zero of one or moreof its linear factors. Therefore, we can find the zerosof a quadratic function by finding the zeros of the simplerlinear functions that are its factors. This means that instead of having to solvea quadratic equation we can solve linear equationswhich are easier. So to find the zeros of a quadratic function,we factor it into linear functions and then identify the zeros of those functions.
But is it always possible to factor a quadraticfunction into linear functions? If the quadratic’s graph intersects the x-axis,the answer is “yes” although when there is only asingle intersection point the two linear factors will be identical However, if we are only considering real numbers it will not be possibleto factor quadratic functions whose graph does not intersect the x-axis. But any quadratic functionwhose graph does intersect the x-axis and therefore has real zeros can always be factored intotwo real linear functions which may either be identical or distinct.
So to factor a quadratic function we must find a pair of linear expressionswhich when multiplied together produce the quadratic expressiondefining the function. Fortunately, there are several easilyrecognized cases of quadratic expressions called “special products” whose factors can be readily identified. In the next two lectures we will examinetwo kinds of special products including quadratic expressionswhich are the “difference of squares” and quadratic expressionswhich are “perfect squares”.