what r the factors of
x^3+13x^2+32x+20
Answers
Answered by
0
Let P(x) =x^3+13x^2+32x+20
If (x - a) is a factor of P(x) then P(a) = 0
So first lets find the a for which P(a) = 0
Lets try the values 1,-1,0,2,-2 for a.
if a = 1
then P(1) = 1^3+13*1^2+32*1+20 = 1 + 13 + 32 + 20 = 66 not equal to zero
so try a = -1
P(-1) = (-1)^3 +13*(-1)^2 + 32*(-1) +20 = -1 + 13 -32 +20 = -33 + 33 = 0
P(-1) = 0 ==> (x--1) = (x+1) is a factor
So divide P(x) by (x+1)
x^2 + 12x +20
x+1 | x^3+13x^2+32x+20
x^3+ x^2
0 +12x^2 +32x
12x^2 +12x
0 +20x + 20
20x + 20
0
therefore P(x) = (x+1)*(x^2+12x+20)
now we need to factor x^2+12x+20
we need to find two numbers whose SUM = 12 and PRODUCT = 20
10 + 2 = 12
and 10* 2 = 20
so split the term 12x in x^2+12x+20 as the sum of 10x and 2x
==> x^2+12x+20 = x^2+10x+2x+20
now x is common for the first 2 terms and 2 is common for the last two.
==> x^2+12x+20 = x^2+10x+2x+20 = x(x+10)+2(x+10)
now x+10 is common
==> x(x+10)+2(x+10) = (x+10) (x+2)
so the factors of P(x) are (x+1), (x+2), (x+10)
If (x - a) is a factor of P(x) then P(a) = 0
So first lets find the a for which P(a) = 0
Lets try the values 1,-1,0,2,-2 for a.
if a = 1
then P(1) = 1^3+13*1^2+32*1+20 = 1 + 13 + 32 + 20 = 66 not equal to zero
so try a = -1
P(-1) = (-1)^3 +13*(-1)^2 + 32*(-1) +20 = -1 + 13 -32 +20 = -33 + 33 = 0
P(-1) = 0 ==> (x--1) = (x+1) is a factor
So divide P(x) by (x+1)
x^2 + 12x +20
x+1 | x^3+13x^2+32x+20
x^3+ x^2
0 +12x^2 +32x
12x^2 +12x
0 +20x + 20
20x + 20
0
therefore P(x) = (x+1)*(x^2+12x+20)
now we need to factor x^2+12x+20
we need to find two numbers whose SUM = 12 and PRODUCT = 20
10 + 2 = 12
and 10* 2 = 20
so split the term 12x in x^2+12x+20 as the sum of 10x and 2x
==> x^2+12x+20 = x^2+10x+2x+20
now x is common for the first 2 terms and 2 is common for the last two.
==> x^2+12x+20 = x^2+10x+2x+20 = x(x+10)+2(x+10)
now x+10 is common
==> x(x+10)+2(x+10) = (x+10) (x+2)
so the factors of P(x) are (x+1), (x+2), (x+10)
Answered by
0
the factors of x^3+13x^2+32x+20 is 3220
prituldave:
i am not sure
Similar questions