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

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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 - 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.

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 ...

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.

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 ...

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 ...

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 ...

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.

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,

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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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 ex

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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 t

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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(

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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 d

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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.

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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

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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 f

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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))/

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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