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?

## Research, Knowledge and Information :

### 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:

## 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)

### 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

### 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)

### 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.

### 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)

### 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.

### 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.

### 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))

### 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))