Guide :

factor the expression 2x^8 + 4x^7 - 6x^6 + 4x^5

If the GCF is 2x^5, what if the expression factored in simplified terms?

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factor the expression 2x^8+4x^7-6x^6+4x^5 - Brainly.com


factor the expression 2x^8+4x^7-6x^6+4x^5 - 2420408. 1. Log in Join now Katie; ... but the answer will be 2x^5 (x^3 + 2x^2 - 3x + 2). Sorry. Comments; Report; 0.
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factor the expression...2x^8+4x^7-6x^6+4x^5 - OpenStudy


factor the expression...2x^8+4x^7-6x^6+4x^5When factoring, always look for a GCF 1st. Is there a GCF here? ... factor the expression...2x^8+4x^7-6x^6+4x^5.
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2x^8+4x^7-6x^6+4x^5= - Get Easy Solution


Simplifying 2x 8 + 4x 7 + -6x 6 + 4x 5 = 0 Reorder the terms: ... Factor out the Greatest Common Factor (GCF), '2x 5 '. 2x 5 (2 + -3x + 2x 2 + x 3) = 0 Ignore the ...
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Greatest common factor and factor expression? ... - OpenStudy


Greatest common factor and factor expression? 2x^8+4x^7-6x^6+4x^5The GCF is 2x^5 Factored it would be: 2x^5(x^3 + 2x^2 - 3x^4 + 2) ... 2x^8+4x^7-6x^6+4x^5.
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FACTOR EACH EXPRESSION, IF IT CANT BE FACTORED ... - Brainly.com


FACTOR EACH EXPRESSION, ... Greatest Common Factor, of 4x and 12. ... it cannot be factored. 6.) 8(4x+3) 7.) 3(2x-3) 8.) 24(1x+2) 9.) 9 ...
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How do you factor the expression 2x^2 - 5x - 3 ? | Socratic


... 2x^2-5x-3 = 2x^2-6x+x-3 = (2x^2-6x)+(x-3 ... 5)^2-49 =(4x-5)^2 - 7^2 =((4x-5)-7)((4x-5)+7) =(4x-12)(4x+2) =(4(x-3))(2(2x+1)) =8(x-3) ... How do you factor the ...
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How do you factor the expression 4x^4-6x^2+2? | Socratic


... 4x^4-6x^2+2 2(2x^4-3x^2+1) 2(2x^2-1)(x^2-1) 2(2x^2-1)(x-1)(x+1) ... How do you factor the expression #4x^4-6x^2+2#? ... How do you factor #2x^2 + 7x ...
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Untitled Document [www.math.brown.edu]


How to factor simple quadratics Ex: Factor x 2 - 4x + 4 ... = x 2 + 4x - 2x - 8 = x 2 + 2x - 8, ... (x 3 - 2x 2 + 4) = x 5 + 2x 4 - 9x 3 + 6x 2 + 16x - 4.
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Factoring in Algebra - Math is Fun


Factoring in Algebra Factors. ... So you can factor the whole expression into: 2y+6 = 2 ... Because 4x 2 is (2x) 2, and 9 is (3) 2,
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Math 20 Sample Questions - Fullerton College


MATH 20 SAMPLE QUESTIONS. ... A. \( 4x^{4} \) B. \( 6x^{8} \) C. \( 3x \) D. \( 12x^{4} \) 8. ... 12. One factor of \( 2x^{2} - 5xy - 3y^{2} \), is:
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Suggested Questions And Answer :


Idot know how to help my soon

u mean 1.   3x10=(2x10)+(1x 10) 2.   2x6 =(2x5)+(2x1) 3.   4x7=(4x5)+(4x2) 4.   11x8=(11x5)+(11x3) 5.   3X6=(3X1)+(3X5) 6.   6X6=(6X2)+(6X4) 7.   7X7=(7X4)+(7X3) 8.   1X8=(1X5)+(1X3)
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how to factorise fully

I can offer tips: Look out for constants and coefficients that are multiples of the same number, e.g., if all the coefficients are even, 2 is a factor. If 3 goes into all the coefficients, 3 is a factor. Place the common numerical factor outside brackets, containing the expression, having divided by the common factor. In 2x^2+10x-6 take out 2 to give 2(x^2+5x-3). In 24x^2-60x+36 12 is a common factor so this becomes: 12(2x^2-5x+3). This also applies to equations: 10x+6y=16 becomes 2(5x+3y)=2(8). In the case of an equation, the common factor can be completely removed: 5x+3y=8. Now look for single variable factors. For example, x^2y^3+x^3y^2. Look at x first. We have x^2 and x^3 and they have the common factor x^2, so we get x^2(y^3+xy^3) because x^2 times x is x^3. Now look at y. We have y^2 and y^3 so now the expression becomes x^2y^2(y+x). If the expression had been 2x^2y^3+6x^3y^2, we also have a coefficient with a common factor so we would get: 2x^2y^2(y+3x). If you break down the factors one at a time instead of all at once you won't get confused. After single factors like numbers and single variables we come to binomial factors (two components). These will usually consist of a variable and a constant or another variable, such as x-1, 2x+3, x-2y, etc. The two components are separated by plus or minus. In (2) above we had a binomial component y+x and y+3x. These are factors. It's not as easy to spot them but there are various tricks you can use to help you find them. More often than not you would be asked to factorise a quadratic expression, where the solution would be the product of two binomial factors. Let's start with two such factors and see what happens when we multiply them. Take (x-1) and (x+3). Multiplication gives x(x+3)-(x+3)=x^2+3x-x-3=x^2+2x-3. We can see that the middle term (the x term) is the result of 3-1, where 3 is the number on the second factor and 1 the number in the first factor. The constant term 3 is the result of multiplying the numbers on the factors. Let's pick a quadratic this time and work backwards to its factors; in other words factorise the quadratic. x^2+2x-48. The constant term 48 is the product of the numbers in the factors, and 2 is the difference between those numbers. Now, let's look at a different quadratic: x^2-11x+10. Again the constant 10 is the product of the numbers in the factors, but 11 is the sum of the numbers this time, not the difference. How do we know whether to use the sum or difference? We look at the sign of the constant. We have +10, so the plus tells us to add the numbers (in this case, 10+1). When the sign is minus we use the difference. So in the example of -48 we know that the product is 48 and the difference is 2. The two numbers we need are 6 and 8 because their product is 48 and difference is 2. It couldn't be 12 and 4, for example, because the difference is 8. What about the signs in the factors? We look at the sign of the middle term. If the sign in front of the constant is plus, then the signs in front of the numbers in the factors are either both positive or both negative. If the sign in front of the constant is negative then the sign in the middle term could be plus or minus, but we know that the signs within each factor are going to be different, one will be plus the other minus. The sign in the middle term tells us to use the same sign in front of the larger of the two numbers in the factors. So for x^2+2x-48, the numbers are 6 and 8 and the sign in front of the larger number 8 is the same as +2x, a plus. The factors are (x+8)(x-6). If it had been -2x the factors would have been (x-8)(x+6). Let's look at a more complicated quadratic: 6x^2+5x-21. (You may also see quadratics like 6x^2+5xy-21y^2, which is dealt with in the same way.) The way to approach this type of problem is to look at the factors of the first and last terms. Just take the numbers 6 and 21 and write down their factors as pairs of numbers: 6=(1,6), (2,3) and 21=(1,21), (3,7), (7,3), (21,1). Note that I haven't included (6,1) and (3,2) as pairs of factors for 6. You'll see why in a minute. Now we make a table (see (5) below). In this table the columns A, B, C and D are the factors of 6 (A times C) and 21 (B times D). The table contains all possible arrangements of factors. Column AD is the product of the "outside factors" in columns A and D and column BC is the product of the "inside factors" B and C. The last column depends on the sign in front of the constant 21. The "twiddles" symbol (~) means positive difference if the sign is minus, and the sum if the sign in front of the constant is plus. So in our example we have -21 so twiddles means the difference, not the sum. Therefore in the table we subtract the smaller number in the columns AD and BC from the larger and write the result in the twiddles column. Now we look at the coefficient of the middle term of the quadratic, which is 5 and we look down the twiddles column for 5. We can see it in row 6 of the figures: 2 3 3 7 are the values of A, B, C and D. If the number hadn't been there we've either missed some factors, or there aren't any (the factors may be irrational). We can now write the factors leaving out the operators that join the binomial operands: (2x 3)(3x 7). One of the signs will be plus and the other minus. Which one is which? The sign of the middle term on our example is plus. We look at the AD and BC numbers and associate the sign with the larger product. We are interested in the signs between A and B and C and D. In this case plus associates with AD because 14 is bigger than 9. The plus sign goes in front of the right-hand operand D and the minus sign in front of right-hand operand B. If the sign had been minus (-5x), minus would have gone in front of D and plus in front of B. So that's it: [(Ax-B)(Cx+D)=](2x-3)(3x+7). (The solution to 6x^2+5xy-21y^2 is similar: (2x-3y)(3x+7y).) [In cases where the coefficient of the middle term of the quadratic appears more than once, as in rows 1 and 7, where 15 is in the twiddles column, then it's correct to pick either of them, because it just means that one of the binomial factors can be factorised further as in (1) above.] Quadratic factors A B C D AD BC AD~BC 1 1 6 21 21 6 15 1 3 6 7 7 18 11 1 7 6 3 3 42 39 1 21 6 1 1 126 125 2 1 3 21 42 3 39 2 3 3 7 14 9 5 2 7 3 3 6 21 15 2 21 3 1 2 63 61  
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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
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Write each rational expression in lowest terms(20 r+10)/(30r+15)

(20r+10)/(30r+15) Factor out the GCF of 10 from each term in the polynomial. (10(2r)+10(1))/(30r+15) Factor out the GCF of 10 from 20r+10. (10(2r+1))/(30r+15) Factor out the GCF of 15 from each term in the polynomial. (10(2r+1))/(15(2r)+15(1)) Factor out the GCF of 15 from 30r+15. (10(2r+1))/(15(2r+1)) Reduce the expression (10(2r+1))/(15(2r+1)) by removing a factor of 5 from the numerator and denominator. (2(2r+1))/(3(2r+1)) Reduce the expression by canceling out the common factor of (2r+1) from the numerator and denominator. (2(2r+1))/(3(2r+1)) Reduce the expression by canceling out the common factor of (2r+1) from the numerator and denominator. (2)/(3)
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k^2-k-20/18

I read this as (k^2-k-20)/18. What pair of factors of 20 differ by 1? This is essentially what the quadratic in k is asking. The pairs of factors of 20 are (1,20), (2,10), (4,5) and the only pair with a difference of 1 is (4,5). The larger of the two has a minus and the other a plus to make the difference -1 which is the coefficient of the middle term. So k^2-k-20 factorises: (k-5)(k+4). And we place this over 18: (k-5)(k+4)/18. This won't factorise any further. When I was reviewing this question I felt there was more to it. Supposing the question was: what values can k take so that the expression is an integer? I spotted something interesting. The zeroes of the quadratic are 5 and -4: either of these values make the expression zero. But 5+4=9, and 9 is a factor of 18, the denominator. We can write the factors of the quadratic as y(y-9)/18, where y replaces k+4; or we can write y(y+9)/18, where y replaces k-5. The pairs of factors of 18 are (1,18), (2,9), (3,6). If the numerator contains any of these pairs, the expression will be an integer, or whole number, positive or negative. What values of y must we have so that y(y-9)/18 is an integer? Let's start with y=0; the expression is zero, which is an integer. Now y=3; we have 3(-6)/18=-1, another integer. y=6; 6(-3)/18=-1. y=9; we get 0. y=12 we get 12*3/18=2. y=15; we get 15*6/18=5. In fact, all multiples of 3 work, so we can write y=3N as the general solution, where N is a positive or negative integer. What about k? We know that y=k+4 or k-5 so k+4=3N or k-5=3N are solutions. We can write these solutions as k=3N-4 or k=3N+5. If N=0 we have -4 and 5, which are the zeroes of the quadratic. If N=1, k=-1 or 8; if N=2, k=2 or 11; if N=3, k=5 or 14; if N=4, k=8 or 17; and so on. If you substitute these and other values of k (according to the formula) you'll see that the expression is always an integer.
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3x³-2x²-19x-6/3x+1

((3x^(3)-2x^(2)-19x-6)/(3x+1)) Factor the polynomial using the rational roots theorem. (((x+(1)/(3))(x+2)(x-3))/(3x+1)) To add fractions, the denominators must be equal.  The denominators can be made equal by finding the least common denominator (LCD).  In this case, the LCD is 3.  Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions. (((x*(3)/(3)+(1)/(3))(x+2)(x-3))/(3x+1)) Complete the multiplication to produce a denominator of 3 in each expression. ((((3x)/(3)+(1)/(3))(x+2)(x-3))/(3x+1)) Combine the numerators of all expressions that have common denominators. ((((3x+1)/(3))(x+2)(x-3))/(3x+1)) Any number raised to the 1st power is the number. ((((1)/(3))(3x+1)(x+2)(x-3))/(3x+1)) Reduce the expression by canceling out the common factor of (3x+1) from the numerator and denominator. (((3x+1)(x+2)(x-3))/((3x+1))*(1)/(3)) Reduce the expression by canceling out the common factor of (3x+1) from the numerator and denominator. ((x+2)(x-3)*(1)/(3)) Multiply the rational expressions to get ((x+2)(x-3))/(3). (((x+2)(x-3))/(3)) Remove the parentheses around the expression ((x+2)(x-3))/(3). ((x+2)(x-3))/(3)
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factorise a5-4a3+8a2-32

The clue is in the number 32 which is 2^5. So a=2 may be a factor. It is, because if we set the expression to zero, a=2 would be a solution. This means (a-2) is a factor, so we can use algebraic long division to simplify the equation. When we do this we get a^4+2a^3+8a+16, a quartic. This time because 2^4=16, we might suspect that a=2 is another solution if the expression were zero. But that would mean another factor of (a-2), but we have no minus signs in the expression. Try a=-2 and the middle terms add up to -32 and the outer terms to +32, so (a+2) is another factor. We divide the quartic by (a+2). This time we're left with a^3+8. Clearly (a+2) divides into this because a=-2 makes the expression zero. Now all we're left with is a^2-2a+4 which factorises into (a-2)^2. So totally factorised we have (a-2)(a+2)(a+2)(a-2)(a-2) or (a+2)^2(a-2)^3.
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a^2b^2 (b-a)+b^2c^2 (c-b)+c^2a^2 (a-c)

Expand the expression: a^2b^3-a^3b^2+b^2c^3-b^3c^2+c^2a^3-c^3a^2. Careful inspection shows that the expression is symmetrical in all three variables. Also, if the expression is equal to zero, putting a=b makes the expression zero, so does b=c and c=a. Therefore (a-b), (b-c) and (c-a) must be factors. Therefore part of the factorisation is (a-b)(b-c)(c-a). But when this is expanded (see below) the sum total of powers in the expansion is 3, but we require a sum equal to 5. For example, we have ab^2 but we need a^2b^3, so we need to multiply by ab. We still need symmetry so we try ab+bc+ac as another factor (see below). (a-b)(b-c)(c-a)(ab+bc+ca)=(ab^2-a^2b+bc^2-b^2c+ca^2-c^2a)(ab+bc+ca)= a^2b^3-a^3b^2+ab^2c^2-ab^3c+a^3bc-a^2bc^2+ ab^3c-a^2b^2c+b^2c^3-b^3c^2+a^2bc^2-abc^3+ a^2b^2c-a^3bc+abc^3-ab^2c^2+a^3c^2-a^2c^3= a^2b^3-a^3b^2+b^2c^3-b^3c^2+a^3c^2-a^2c^3. This matches the original expression, because all the other terms cancel out.
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4a^2 + 1/(4a^2) -2 +4a -1/(4a)

(4a^(2)+1)/(4a^(2))-2+(4a-1)/(4a) Multiply each term by a factor of 1 that will equate all the denominators.  In this case, all terms need a denominator of 4a^(2). The ((4a-1))/(4a) expression needs to be multiplied by ((a))/((a)) to make the denominator 4a^(2). (4a^(2)+1)/(4a^(2))-2*(4a^(2))/(4a^(2))+(4a-1)/(4a)*(a)/(a) Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 4a^(2). (4a^(2)+1)/(4a^(2))-(2*4a^(2))/(4a^(2))+(4a-1)/(4a)*(a)/(a) Multiply 2 by 4a^(2) to get 8a^(2). (4a^(2)+1)/(4a^(2))-(8a^(2))/(4a^(2))+(4a-1)/(4a)*(a)/(a) Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 4a^(2). (4a^(2)+1)/(4a^(2))-(8a^(2))/(4a^(2))+((4a-1)(a))/(4a^(2)) The numerators of expressions that have equal denominators can be combined.  In this case, ((4a^(2)+1))/(4a^(2)) and -((8a^(2)))/(4a^(2)) have the same denominator of 4a^(2), so the numerators can be combined. ((4a^(2)+1)-(8a^(2))+((4a-1)(a)))/(4a^(2)) Simplify the numerator of the expression. (-a+1)/(4a^(2))
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Express f(z) in the form f(z) = (z + a)(z^2 + bz + c)

QUESTION: Given that (z + 2 - 3i) is a factor of f(z) = z^3 + 9z^2 + 33z + 65, Express f(z) in the form f(z) = (z + a)(z^2 + bz + c), Where a, b and c are integers. Since a and c are integers, then an expansion of f(z) = (z + a)(z^2 + bz + c) would show that a*c = 65. The only factors of 65 are 5 and 13. So a = 5 and c = 13, or vice versa. Expanding f(z) = (z + a)(z^2 + bz + c) we get, f(z) = z^3 + (b+a)z^2 + (c+ab)z + ac comparing coefficients with f(z) = z^3 + 9z^2 + 33z + 65, this gives us, b + a = 9 c + ab = 33 ac = 65 Test: if a = 5 b + a = 9 ---> b = 4 ac = 65 ---> c = 13 c + ab = 33 ---> a = 5 (this verifies that a equal 5) a = 5, b = 4, c = 13 Then f(z) = z^3 + 9z^2 + 33z + 65 = (z + 5)(z^2 + 4z + 13) BTW: if you use (z + 2 - 3i) as a factor. you end up with f(z) = (z + 2 - 3i)(z^2 + (7+3i)z + (10+15i))
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