Guide :

# if (x+1) is a factor of ax cube+x square-2x+4a-9 find a

polynomials

## Research, Knowledge and Information :

### IF (x+1) IS A FACTOR OF ax3+x2-2x+4a-9 , FAND THE VALUE OF ...

... (x+1) IS A FACTOR OF ax3+x2-2x+4a-9 ... = ax 3 +x 2-2x+(4a-9). It is given that (x+1) ... what is the formula of diagonal of cube and cuboid.

### if x+ 1 is a factor of ax power 3 + x power 2 - 2x + 4a - 9 ...

if x+ 1 is a factor of ax power 3 + x power 2 - 2x + 4a - 9 find the value of ' a ' - 1213161

### The polynomial [math]x^3-3x^2+4x-1[/math] is a factor of ...

... x^3-3x^2+4x-1[/math] is a factor of [math]x^9+px^6+qx^3+r[/math]. What is ... (x^3 - 3x^2 + 4x - 1)(x^6 + ax^5 + bx^4 + cx^3 ... x^6[/math]: [math]-1 + 4a -3b + c ...

### If x+a is a factor of x^3+ax^2-2x+a+4 then a equals? - Quora

Thanks for A2A. This problem deals with many diverse applications in higher algebra. Since x+a is a factor of given polynomial (f(x)), which implies polynomi...

### if x-2 is a factor of polynomial x cube - 2ax square + ax - 1 ...

if x-2 is a factor of polynomial x cube - 2ax square + ax - 1 what is the ... if x-2 is a factor of polynomial x cube - 2ax square + ax ... (-4a squared + b cube ...

### 7 Quadratic Expressions - UNE - University of New England

in ax.) In (x), the square of x ... Find the value of f(x) = 2x2 +4x−9 when (a) x ... Add b2 to both sides to turn the left-hand side into a perfect square i.e. 4a ...

### Factor Polynomials - analyzemath.com

1: One common factor. a x + a y = a ... 9: Perfet square x 2 - 2xy + y 2 = (x - y) 2 10: Perfect cube x 3 + 3x 2 y + 3xy 2 + y 3 = ...

### Factoring Polynomials - Ashworth College

(x – 2)(6x – 2) Common factor (2x – 1)(3x – 4) ... x2 + 6x +9 is a perfect-square trinomial because ... Factoring Polynomials

### Factoring trinomials in the form of ax^2+bx+c | StudyPug

Try these guided questions to learn how to factor trionomials in the form of ax ... how to factor trinomials. Question 1. ... + b x + c. 9. Factoring perfect square ...

## Suggested Questions And Answer :

### find the length of each boundary

find the length of each boundary Vista county is setting aside a large parcel of land to preserve it as open space. the county has hired Jane's surveying firm to survey the parcel, which is in the shape of a right triangle. the longer leg of the triangle measures 5 miles less than the square of the shorter leg, and the hypotenuse of the triangle measures 13 miles less than twice the square of the shorter leg. the length of each boundary is a whole number. find the length of each boundary. Make a the long side, b the short side and c the hypotenuse. a = b^2 - 5 b = b, obviously c = 2b^2 - 13 From the Pythagorean theorem, we have a^2 + b^2 = c^2 (b^2 - 5)^2 + b^2 = (2b^2 - 13)^2 (b^2 - 5) * (b^2 - 5) + b^2 = (2b^2 - 13) * (2b^2 - 13) b^4 - 5b^2 - 5b^2 + 25 + b^2 = 4b^4 - 26b^2 - 26b^2 + 169 b^4 - 9b^2 + 25 = 4b^4 - 52b^2 + 169 Multiply by -1. That way, when the terms on the right side are moved to the left side, the b^4 coefficient will be positive. -b^4 + 9b^2 - 25 = -4b^4 + 52b^2 - 169 -b^4 + 4b^4 + 9b^2 - 52b^2 - 25 + 169 = 0 3b^4 - 43b^2 + 144 = 0 We need to factor the left side. If we use 12 and 12 as the factors for 144, we don't get the required coefficient for the b^2 term. The factors 16 and 9 will work because we multiply the 9 by 3. (3b^2 - 16)(b^2 - 9) = 0 Using those factors we have (3b^2 -16) = 0 or (b^2 - 9) = 0 Solving for b in the first factor: (3b^2 -16) = 0 3b^2 = 16 b^2 = 16/3 = 5.33333 b = 2.3094 This doesn't satisfy the requirement that all lengths are whole numbers. Solving for b in the second factor: (b^2 - 9) = 0 b^2 = 9 b = 3 Actually, b = -3 is also valid, but the length cannot be negative. We have side b = 3 miles, a = b^2 - 5 = 9 - 5 = 4 miles, c = 2b^2 - 13 = 2(9) - 13 = 18 - 13 = 5 miles So, we have a simple 3-4-5 right triangle.

### how do you use break apart to divide ?

You need to find the factors of the number you're dividing. If it's an even number you can divide by 2, so do this and note the factors as you find them. Therefore you have so far 2 and another number, we'll call n. Look at n, and if it's even you have another 2, so you would write 2*2 and another n. When you no longer have an even number, or if the first number was odd, you check to see if 3 goes into it. You make a note and divide the number by 3 so you now have ...3 and another n. You may even have 2*2...*3*n. Again you try 3, until you can't divide exactly, then you move on to 5 (numbers ending in 5 or 0 are exactly divisible by 5). You carry on like this noting what factors you've found as you go along. The numbers you pick as divisors will always be prime: in order here are the first few: 2, 3, 5, 7, 11, 13. A prime number is one which can't be divided exactly by a number smaller than it (actually up to its square root). 4 isn't prime because 2 goes into it; 9 isn't prime because 3 goes into it and all primes apart from 2 are odd numbers.  Now's a good time to look at examples. First, 660. It's even so 2 goes into it. So we have 2*330. 330 is also even: 2*2*165. 165 is divisible by 3, so we divide by 3 and we have 2*2*3*55. 55 ends in 5 and it's not divisible by 3, so we divide by 5 and get 11, and now we have 2*2*3*5*11. 11 is prime so we stop. 660=2*2*3*5*11. This breakdown means we can spot the factors other than prime factors. 2*2 means that 4 is a factor; 2*3 means that 6 is a factor; 3*5 means that 15 is a factor; 2*5*11=110 means that 110 is a factor, and so on. So if you needed to divide 660 by 15 you cross out the factors that make 15 (3 and 5) leaving 2*2*11=44. Let's try another example: 2030. 2030=2*1015=2*5*203=2*5*7*29. 29 is prime. How do we know how far to divide? Take 9042=2*4521=2*3*11*137. Is 137 a prime number? We only have to try and divide by numbers up to the square root of 137. What does that mean? The square root is the number which when multiplied by itself is closest to the number we're trying to divide, but smaller than it. 11*11=121; 12*12=144, so we only have to go up to dividing by 11, because 11*11 is less than 137 while 12*12 is greater, and the square of the next prime is 13*13=169, which is bigger than 137; so we only need to go up to primes up to 11, which doesn't divide exactly into 137, so 137 must be a prime number. Now try some numbers yourself!

### Factoring perfect square trinomials

Numerically, a perfect square is a number that is the product of another number multiplied with itself ( 4 x 4 = 16.  16 is a perfect square). For trinomials, a perfect square trinomial is a trinomial that is the product of a binomial multiplied with itself  (   (2x-5)(2x-5) = 4x^2 - 20x + 25.  4x^2 - 20x + 25 is a perfect square trinomial). If you know a trinomial is a perfect square going into the problem, factoring it is relatively easy.  Just find the square root of the lead term, the square root of the last term, and decide whether a + or - sign belongs in the middle. For example,  in 4x^2 - 20x + 25,  the square root of 4x^2 is 2x.  The square root of 25 is 5.  My solution will either be (2x+5)(2x+5)   or   (2x-5)(2x-5).  FOILing these out (as you should always do to check your answer) we find that the minus sign works. 4x^2 - 20x + 25 = (2x - 5)^2

### Find the exact solutions of each system of equations 4x+y^2=20 4x^2+y^2=100

Find the exact solutions of each system of equations 4x+y^2=20 4x^2+y^2=100 We need to find the exact solutions of each system of equations written as ordered pairs. 1) 4x+y^2=20 2) 4x^2+y^2=100 Subtract equation one from equation two, eliminating y^2.   4x^2    + y^2 = 100 -(     4x + y^2 =  20) ------------------------   4x^2 - 4x     = 80 4x^2 - 4x = 80 3) 4x^2 - 4x - 80 = 0 Solve equation three for x. Factor the left side. 4x^2 - 4x - 80 = 0 (4x - 20)(x + 4) = 0 One or both factors must equal zero. Factor one: (4x - 20) = 0 4x = 20 x = 5 Factor two: (x + 4) = 0 x = -4 Plug both of those values into equation one to solve for y. 4x + y^2 = 20 4(5) + y^2 = 20 20 + y^2 = 20 y^2 = 0 y = 0 (5, 0) 4x + y^2 = 20 4(-4) + y^2 = 20 -16 + y^2 = 20 y^2 = 20 + 16 y^2 = 36 y = ±6     (you can get 36 by squaring 6 and by squaring negative 6) (-4, -6) (-4, 6) Use equation two to verify the various values. (5, 0) 4x^2 + y^2 = 100 4(5^2) + 0^2 = 100 4(25) + 0 = 100 100 = 100 (-4, -6) 4x^2 + y^2 = 100 4(-4^2) + (-6^2) = 100 4(16) + (36) = 100 64 + 36 = 100 100 = 100 (-4, 6) 4x^2 + y^2 = 100 4(-4^2) + 6^2 = 100 4(16) + 36 = 100 64 + 36 = 100 100 = 100 All three pairs are valid: (5, 0), (-4, -6) and (-4, 6)

### The pythagorean theorem of this problem x^2+(2x+6)^2=(2x+4)^2

Pythagoras' theorem relates the lengths of the sides of a right-angled triangle: a^2+b^2=c^2 where c, the hypotenuse, is the longest side. In your question 2x+6 is the longest side, so there could be no solution, since the the sum of the two squares of the sides on the left-hand side must be bigger than the square of a side that's smaller than either one of them, and 2x+4 is smaller than 2x+6 (2x+6-(2x+4)=2). The equation x^2+(2x+4)^2=(2x+6)^2 can be solved. We can write it as: x^2=(2x+6)^2-(2x+4)^2. On the right-hand side we have the difference of two squares, which factorises to make the calculation simpler: (2x+6-(2x+4))(2x+6+2x+4)=2(4x+10). So now we have x^2=8x+20, by expanding the brackets. We can write this: x^2-8x=20 by subtracting 8x from each side. This is a quadratic that can be solved by completing the square. Halve the x coefficient then square it: 8/2=4, and 4^2=16. Add 16 to both sides: x^2-8x+16=36. The left-hand side is a perfect square of x-4: (x-4)^2=36. 36 is also a perfect square=6^2 so, by taking square roots of each side we have: x-4=6 or -6, because both 6 and -6 have a square of 36. We have two solutions: x=4+6=10 or 4-6=-2. Another way of solving for x is factorisation: x^2-8x=20 can also be written: x^2-8x-20=0 which factorises: (x-10)(x+2)=0, so either x-10=0 or x+2=0, and that also gives us x=10 and -2. The factors x-10 and x+2 are binomial expressions, and the equations arising from them use the zero factor idea: x-10=0 or x+2=0 to solve for x.

### sqrt(3a+10) = sqrt(2a-1) + 2

sqrt(3a+10)=sqrt(2a-1)+2 To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(3a+10))^(2)=(~(2a-1)+2)^(2) Simplify the left-hand side of the equation. 3a+10=(~(2a-1)+2)^(2) Squaring an expression is the same as multiplying the expression by itself 2 times. 3a+10=(~(2a-1)+2)(~(2a-1)+2) Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group. 3a+10=(~(2a-1)*~(2a-1)+~(2a-1)*2+2*~(2a-1)+2*2) Simplify the FOIL expression by multiplying and combining all like terms. 3a+10=(~(2a-1)^(2)+4~(2a-1)+4) Remove the parentheses around the expression ~(2a-1)^(2)+4~(2a-1)+4. 3a+10=~(2a-1)^(2)+4~(2a-1)+4 Raising a square root to the square power results in the expression inside the root. 3a+10=(2a-1)+4~(2a-1)+4 Add 4 to -1 to get 3. 3a+10=2a+3+4~(2a-1) Since a is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. 2a+3+4~(2a-1)=3a+10 Move all terms not containing ~(2a-1) to the right-hand side of the equation. 4~(2a-1)=-2a-3+3a+10 Simplify the right-hand side of the equation. 4~(2a-1)=a+7 Divide each term in the equation by 4. (4~(2a-1))/(4)=(a)/(4)+(7)/(4) Simplify the left-hand side of the equation by canceling the common terms. ~(2a-1)=(a)/(4)+(7)/(4) To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(2a-1))^(2)=((a)/(4)+(7)/(4))^(2) Simplify the left-hand side of the equation. 2a-1=((a)/(4)+(7)/(4))^(2) Combine the numerators of all expressions that have common denominators. 2a-1=((a+7)/(4))^(2) Expand the exponent of 2 to the inside factor (a+7). 2a-1=((a+7)^(2))/((4)^(2)) Expand the exponent 2 to 4. 2a-1=((a+7)^(2))/(4^(2)) Simplify the exponents of 4^(2). 2a-1=((a+7)^(2))/(16) Multiply each term in the equation by 16. 2a*16-1*16=((a+7)^(2))/(16)*16 Simplify the left-hand side of the equation by multiplying out all the terms. 32a-16=((a+7)^(2))/(16)*16 Simplify the right-hand side of the equation by simplifying each term. 32a-16=(a+7)^(2) Since (a+7)^(2) contains the variable to solve for, move it to the left-hand side of the equation by subtracting (a+7)^(2) from both sides. 32a-16-(a+7)^(2)=0 Squaring an expression is the same as multiplying the expression by itself 2 times. 32a-16-((a+7)(a+7))=0 Multiply -1 by each term inside the parentheses. 32a-16-a^(2)-14a-49=0 Since 32a and -14a are like terms, add -14a to 32a to get 18a. 18a-16-a^(2)-49=0 Subtract 49 from -16 to get -65. 18a-65-a^(2)=0 Move all terms not containing a to the right-hand side of the equation. -a^(2)+18a-65=0 Multiply each term in the equation by -1. a^(2)-18a+65=0 For a polynomial of the form x^(2)+bx+c, find two factors of c (65) that add up to b (-18).  In this problem -5*-13=65 and -5-13=-18, so insert -5 as the right hand term of one factor and -13 as the right-hand term of the other factor. (a-5)(a-13)=0 Set each of the factors of the left-hand side of the equation equal to 0. a-5=0_a-13=0 Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides. a=5_a-13=0 Set each of the factors of the left-hand side of the equation equal to 0. a=5_a-13=0 Since -13 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 13 to both sides. a=5_a=13 The complete solution is the set of the individual solutions. a=5,13

### Finding the nontrivial perfect square of a given quadratic form?

assuming z^2+180-88=m^2 rewrite it in a form of z(z+180) = m^2 + 88 since LHS is a form of 2 factors z and z+180, the right handside must be also in such a form as a x b therefore m must have same factors as 88, you can start from there and try the factors of 88

### find the x and y intercepts - y= x2 (square) - x - 6

to find y intercept, let x=0.  So -y = (0)^2 - 0 - 6, -y = -6 so y=6. to find x intercept let y=0.  So, 0=x^2 - x - 6.  This trinomial is factorable into 0=(x-3)(x+2) let each factor=0, then x-3=0 and x+2=0, solve each for x and intercepts are x=3 and x=(-2)

### factor the polynomial x2+9x+20

Problem: factor the polynomial x2+9x+20 i need to find the answer to thos qustion ,this question is pre-algebra ,i need to know how to factor ploynomials,what is a polynomial♦ A polynomial is an expression (as given in your statement) with one or more unknowns (in this case, x), co-efficients (the numbers that multiply the unknowns) and positive exponents (the 2 in x squared). Find factors of the constant (20) which add up to the co-efficient of x (9). x^2 + 9x + 20 = (x + 5) * (x + 4)

### How do special products help us factor polynomials? Give examples.

The special products that students are usually asked to identify and remember to help in factorisation are the squares and difference of squares: (a+b)^2=a^2+2ab+b^2; (a-b)^2=(b-a)^2=a^2-2ab+b^2; (a-b)(a+b)=a^2-b^2. a and b can be composite quantities like 3x, xy, 2xyz, etc. Examples: (3xy-z)^2=9x^2y^2-6xyz+z^2; (1/x+1/y)^2=1/x^2+2/xy+1/y^2; (5-t)(5+t)=25-t^2. I include some other types of products you might find useful: The constant term in a polynomial is a strong clue to how it factors, assuming that it is meant do so. If the constant is a prime number p, then the factors will consist of p and at least one 1, because p*1*...*1= p. The degree of the polynomial tells you how many factors there are: a degree 2 (quadratic) has two, for example. Examples: x^3-7x^2-x+7=(x-1)(x+1)(x-7); x^2-12x-13=(x-13)(x+1). If the constant is a composite number (i.e., not a prime number), then you need to factorise it and group the factors according to the degree. For example, if the constant is 15 and the degree is 4, then the factors arranged in fours are (1,1,1,15), (1,1,3,5). Example: x^4+2x^3-16x^2-2x+15=(x-1)(x+1)(x-3)(x+5). If the degree is n and the constant term is p^n, where p is a prime number, then the polynomial may consist of factors +p and -p. Examples: x^3-5x^2-25x+125=(x-5)^2(x+5). But x^2-26x+25=(x-25)(x-1). The sign of the constant is significant. If the degree n is even, and the sign is plus, then the zeroes of the polynomial will consist of an even number of pluses and minuses. If the sign is minus, there is an odd number of pluses and an odd number of minuses. If the degree is odd, and the sign is plus, the factors contain an even number of minuses, or none at all, and the rest are plus. If the sign is minus the factors contain an odd number of minuses, and the rest are pluses, or there are no other factors.