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x^2-x-20=0 factor please? is the answer (x-5)(x+4) its been awhile

I can't remember how to factor. Do I need to use quadratic formula for this or can I just easily factor it? What method should I use to solve this? I don't remember how to do basic factoring...

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Algebra II Final Flashcards | Quizlet


x + 5 Mathway (Algebra) Factor. ... (maybe 0.315 or 32%) Solution.11 + .20 + .31 + 30 + .15 + 24 = 1 (or ... Find h when t = 5.5 s. Round your answer to the nearest ...
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Factoring Quadratics: The Simple Case (page 1 of 4) - Purplemath


Factoring Quadratics: The Simple Case ... Factor x 2 + 5x + 6. ... This is the answer: x 2 + 5x + 6 = (x + 2)(x + 3)
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Help! 14 points A dilation with center (0,0) and scale factor ...


... (0,0) and scale factor 0.5 maps (4, -10) to ... (10,-5.4) is the answer Comments; Report; 2 5 2. ... PLEASE!!!! consider triangle PQR what is the length side of QR?
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11. Factoring and solving equations - Wellesley College


Findthat x-I gives 13+3.12-4 -0. So x-1 isa factor of x3 + 3x2 - 4 . ... Exam& Solve x2-2x-3=0 for x. Method: Factoring. x2 - 2x - 3 = ... 4. x2 - (x-2)2x = 4 5. x=-4 ...
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Solving Quadratic Equations: Examples - Purplemath | Home


Solving Quadratic Equations: Examples ... Solve x 2 + x – 4 = 0. ... which means the quadratic could have been factored. But I've got my answer, ...
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Simplifying Rational Expressions - montereyinstitute.org


When x = 4, the denominator is equal to 0. Answer . ... the values of x and y that make the numerator equal to 0. Factor the ... x 3 – x 2 – 20x = 0 . x(x 2 – x ...
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9th Grade Homework Help - jiskha.com


please solve for either x or y: x/2 + y/4 = 7 ... 9th grade whats the answer for xsquare+12x+20=0 ... How do you factor this expression? 10x+15
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20D - Homework Assignment 1 - UCSD Mathematics | Home


MATH 20D Homework Assignment 1 ... y0= x2=y (2) y0= x2=y(1 + x3) (3) y0+ y2 sinx= 0 (4) y0= (3x2 1)=(3 + 2y) ... answer given in the back of the book.)
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Java 170 Midterm Flashcards | Quizlet


Java 170 Midterm. OCC Spring 2011. ... Make sure that your answer is the correct type: 2 is an int 2.0 is a double ... 53 / 5 / (0.6 + 1.4) / 2 + 13 / 2. 8.5.
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Suggested Questions And Answer :


(2x+1)(3x-1)=0

(2x+1)(3x-1)=0 can u please show me the steps to solve this problem thanks The first impulse would be to multiply those quantities, but that would be a waste of time, because you would then need to factor the result. You already have the factors staring you in the face. What the problem is telling you is that one, or both, of those factors has to be equal to zero. Remember, multiplying by zero gives a zero result. So, we set each quantity to zero and solve for x. 2x + 1 = 0 2x = -1 x = -1/2 3x - 1 = 0 3x = 1 x = 1/3 Now, if you want to, you can multiply the factors and use the resulting equation to check your answers. (2x + 1)(3x - 1) = 0 6x^2 + 3x - 2x - 1 = 0 6x^2 + x - 1 = 0 Plug in the values calculated for x. 6x^2 + x - 1 = 0 6(-1/2)^2 + (-1/2) - 1 = 0 6(1/4) - 1/2 - 1 = 0 1 1/2 - 1/2 - 1 = 0 0 = 0 That one checks. 6x^2 + x - 1 = 0 6(1/3)^2 + 1/3 - 1 = 0 6(1/9) + 1/3 - 1 = 0 2/3 + 1/3 - 1 = 0 0 = 0 That one checks, too. x = -1/2   and  x = 1/3
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2x^3+16x^2+32x=0

4x^2-32x+16=0
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solve x²+7x+12= 0

solve x²+7x+12= 0 I need to know how to work it out - please help!!!!! x² + 7x + 12 = 0 We need to find factors of 12 that add up to 7. They must be 3 and 4 (x + 3)(x + 4) = 0 One, or both, of those factors on the left must equal zero. (Multiplying by zero produces a zero result.) Set each one to zero and solve for x. (x + 3) = 0 x = -3 (x + 4) = 0 x = -4 Check both answers using the original equation. x² + 7x + 12 = 0 -3² + 7(-3) + 12 = 0 9 - 21 + 12 = 0 0 = 0 x² + 7x + 12 = 0 -4² + 7(-4) + 12 = 0 16 - 28 + 12 = 0 0 = 0 So, x = -3 and x = -4 are solutions. If you graph this equation, you will get a parabola that crosses the x axis at two points: (-3, 0) and (-4, 0)  
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(2x^2+x-6)/(x+2)=

2x^2+x-6 factorises: (x+2)(2x-3) so we have (x+2)(2x-3)/(x+2). There is a common factor in the denominator and numerator: x+2, so divide them by this common factor: 2x-3 is the answer. There's just one thing: you can only divide through by a common factor if the factor isn't zero, in other words x+2 cannot be zero, so x cannot be -2. Any other value except x=-2 is OK. This is because we can't divide by zero. How did I know how to factorise the top? Well, I guessed that the denominator was likely to be a factor, so I used the "solution" of x+2=0, subtracting 2 from each side: x=-2, then I substituted this value of x into 2x^2+x-6: 2*(-2)*(-2)-2-6=8-2-6=0, which tells me that x+2 divides into the numerator. To find the other factor making up the quadratic, I asked myself what do I need to multiply x by to get 2x^2, and the answer is 2x^2/x=2x; then I asked, what do I need to multiply 2 by to get 6? 6/2=3. Finally, what sign + or - is needed to go in front of 3? We already have +2 in the factor x+2, so we need a minus to make -6. So the other factor of the quadratic must be 2x-3.
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f(x)=x(x+2)(3x-3)/(x-4)(x+3) find x-int,y-int,vertical,horizontal asymotote

Fortunately the function has been factorised for us, so we can see what values of x in and outside the bracketed factors are needed to make those factors zero. That's a good start for answering the question, which appears to be a long-standing one. The values of x that produce zero factors are: 0, -2, 1, 4 and -3. These, as we'll see, have different meanings, depending on where they are in the function. The factors in the numerator simply make the function zero; but those in the denominator give rise to vertical asymptotes. The vertical asymptotes are therefore the lines x=4 and x=-3. The other three values of x represent intercepts: y=f(x)=0 when x=0, -2 and 1. So the x intercepts are (0,0), (-2,0) and (1,0). When x=0, f(x)=0, so the origin (0,0) is both an x and a y intercept. There are no others. There is another asymptote which is neither horizontal nor vertical. When x is very, very, large, the numbers become insignificantly small by comparison, and the function becomes 3x^3/x^2=3x. The line y=3x is therefore an asymptote. This is a line with positive slope 3 passing through the origin. The graph, which is in three parts, never quite touches this line. The central portion sits between the two vertical asymptotes, while the other two sections are outside the asymptotes and these sections are the ones to which the asymptote y=3x applies as well as the vertical asymptotes.
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Write the polynomial in factored form. x^3 + 7^2 + 15x + 9

If you mean x^3 + 7^2 + 15x + 9 then: x^3 + 49 + 15x + 9 x^3 + 15x + 58 does not have any rational zeros. (can't factor) If you mean x^3 + x^2 + 15x + 9 then: . . .it also appears to not have any rational zeros. If you mean x^3 + 7x^2 + 15x + 9 then: Let's try -1. . . f(x) = x^3 + 7x^2 + 15x + 9 f(-1) = -1 + 7 - 15 + 9 f(-1) = 0 So -1 should work as a root, meaning we should be able to factor out (x+1). Note:  If -1 is a root then x = -1 means x + 1 = 0, which means (x+1) should be a factor in x^3 + 7x^2 + 15x + 9 (x + 1)(???x^2 + ??x + ?) = x^3 + 7x^2 + 15x + 9 x * ???x^2 = x^3, so ??? = 1 1 * ? = 9, so ? = 9 (x + 1)(x^2 + ??x + 9) = x^3 + 7x^2 + 15x + 9 The 15x comes from x * 9 + ??x * 1: 9x + ??x = 15x ??x = 6x ?? = 6 (x + 1)(x^2 + 6x + 9) = x^3 + 7x^2 + 15x + 9 (x + 1)(x + 3)(x + 3) Answer:  (x+1)(x+3)^2
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x^2-x-20=0 factor please? is the answer (x-5)(x+4) its been awhile

x^2-x-20=0  if X^2  + aX + b = 0 then ; product of roots is b sum of roots is a there you have to look for numbers whose sum is a and product is b in the equation we have x^2-x-20=0 it can be easily seen that -5 + 4 = -1 -5 * 4 = -20 then we shall have to expand  the equation we have X^2 +4X -5X -20 = 0  X(X+4) -5(X+4) = 0    from (X^2 +4X) +(-5X -20) then (X-5)(X+4)=0
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factor using the trial-and-error method: 12x squared - 5x -3

We need to be careful with this function, because it's quadratic and therefore it could have two roots, two solutions for x when the function=0. These solutions determine the factors. If we plot the graph of the function roughly we can see that when x is large and positive or negative, the function is positive. If we put x=0 we can see where the graph crosses the vertical axis, -3 in this case. To get to -3 the graph must intersect the x-axis at two points. The graph has a U shape. Let's put a couple of values into the function. At x=1 the function is 4. At x=-1 it's 14. So we know it cuts the x-axis twice between x=1 and -1. That gives us a start. Using a calculator to speed things up, we can try values in these limits. Let's go for x=0.5 and -0.5. The two values are respectively are -2.5 and 2.5. What does that tell us? Between x=0 and 0.5 the function goes from -3 to -2.5, and between -0.5 and 0 it goes from -3 to 2.5. So in one case the graph doesn't cross the x-axis but in the other case it does because we've gone from negative to positive between x=0 and -0.5. We need to look first for the solution between -0.5 and 0, so we can try values between -0.4 and -0.1. When we do this we find we go from positive to negative between -0.4 and -0.3. So using trial and error we continue to the next decimal place and try values between -0.39 and -0.31. This time the sign change takes place between -0.34 and -0.33. This is beginning to look like -1/3. Let's put this value into the function. Yes! x=-1/3 is a value that makes the function zero. Therefore one factor is (3x+1). The other factor can be found by inspection. It must start with 4x to make 12x^2 in the function, and the numerical part must be -3 because we have -3 in the original function as the last term and -3*1=-3. So factorisation gives us (3x+1)(4x-3).
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i need help with these questions please and can u explain how u got the answers to please??

x^2-x-4=0 me yuze Quadratik equashun zeroes: 2.5615528 & -1.5615528 a=1, b=-1, c=-4 b^2-4ac=17, so sqrt(17)=4.123105626 -b/2a=0.5 root/2a=2.061552812808 tell teech tu giv a problem weer anser be sumthun simpel, like 2
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factor 2x^3-2x^2-12

factor 2x^3-2x^2-12   I used syntetic division to find a factor.  remember that there is a 0x in the equation 2 |  2     -2     0     -12    |         -4    12     12    |-----------------------------       2     -6    12      0 (x + 2)(2x^2 -6x + 12) (x+2)2(x^2 - 3x + 6) use quadratic equation to finish
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