1. 19-3r+r^2 factor the equation
2. a^2-2ax-80x^2 factor the trinomial
3. c^2+11c+18 factor the trinomial
1. 19-3r+r^2 factor the equation?
2009-07-15 6:00 pm
回答 (5)
2009-07-15 6:12 pm
✔ 最佳答案
1. Impossible! 19 is prime, so we would have to use 1 & 19 to somehow get 3.2. (a+8x)(a-10x)
3. (c+9)(c+2)
;)
2016-12-25 2:07 am
multiply by technique of 9 to do away with fractions =(9r^2 + 6r + a million)/9 =((3r+a million)^2) / 9 through fact the bracket is squared, you may divide it by technique of regardless of squared = 9, that's 3 (taking the beneficial answer) providing you with (r + a million/3)^2
2009-07-15 6:48 pm
1)
19 - 3r + r^2 (this is not an equation. It is an expression or a trinomial)
= r^2 - 3r + 19
(it cannot be factored)
2)
a^2 - 2ax - 80x^2
= a^2 + 8ax - 10ax - 80x^2
= (a^2 + 8ax) - (10ax + 80x^2)
= a(a + 8x) + 10x(a + 8x)
= (a + 8x)(a + 10x)
3)
c^2 + 11c + 18
= c^2 + 9c + 2c + 18
= (c^2 + 9c) + (2c + 18)
= c(c + 9) + 2(c + 9)
= (c + 9)(c + 2)
19 - 3r + r^2 (this is not an equation. It is an expression or a trinomial)
= r^2 - 3r + 19
(it cannot be factored)
2)
a^2 - 2ax - 80x^2
= a^2 + 8ax - 10ax - 80x^2
= (a^2 + 8ax) - (10ax + 80x^2)
= a(a + 8x) + 10x(a + 8x)
= (a + 8x)(a + 10x)
3)
c^2 + 11c + 18
= c^2 + 9c + 2c + 18
= (c^2 + 9c) + (2c + 18)
= c(c + 9) + 2(c + 9)
= (c + 9)(c + 2)
2009-07-15 6:25 pm
1) the roots of (r^2 -3r + 19) are: (3+ sqrt(9-76))/2 & (3- sqrt(9-76))/2
=> 1.5 + i (sqrt(69)/2) & 1.5 - i (sqrt(69)/2)
let, 1.5 + i (sqrt(69)/2) =a
& 1.5 - i (sqrt(69)/2) =b
so, the factors are: (r-a) and (r-b)
(2),(3) already well answered
=> 1.5 + i (sqrt(69)/2) & 1.5 - i (sqrt(69)/2)
let, 1.5 + i (sqrt(69)/2) =a
& 1.5 - i (sqrt(69)/2) =b
so, the factors are: (r-a) and (r-b)
(2),(3) already well answered
2009-07-15 6:12 pm
1. 19-3r+r^2
cannot be factored.
3)
c^2+11c+18
c^2+9c+2c+18
c(c+9)+2(c+9)
(c+9)(c+2)
2)
a^2-10ax+8ax+80
a(a-10x)+8x(a-10x)
(a-10x)(a+8x)
cannot be factored.
3)
c^2+11c+18
c^2+9c+2c+18
c(c+9)+2(c+9)
(c+9)(c+2)
2)
a^2-10ax+8ax+80
a(a-10x)+8x(a-10x)
(a-10x)(a+8x)
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