factorise:x^3+13x^2+32x+20
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Answered by
1
IN THE ATTACHMENT!!
MARK BRAINLIEST!!
u have to do it with hit and trial method!!
if the coefficient of x^3 is there then multiply it with the constant and then find it factors. After that do hit and trial method. if P(x) = (comes) 0 then that is a factor of polynomial.
MARK BRAINLIEST!!
u have to do it with hit and trial method!!
if the coefficient of x^3 is there then multiply it with the constant and then find it factors. After that do hit and trial method. if P(x) = (comes) 0 then that is a factor of polynomial.
Attachments:
manash7:
but we can do it by this way also x^3+13x^2+32x+20=x^3+x^2+12x^2+12x+20x+20=x^2(x+1)+12x(x+1)+20(x+1)=(x^2+12x+10)(x+1)=x^2+10x+2+20)(x+1)=>x(x+10)+2(x+10)(x+1)=>(x+2)(x+10)(x+1)
ohh ya
Answered by
1
Use my simple trick,
It will have 3 roots at it is a cubic polynomial,
Now guess those 3 such that their product satisfies the constant term and
their sum satisfies the coefficient of variable with degree 2.
Now, by the method of finding possible roots we can say,
±( 1,2,4,5,10,20) are it's roots.
Now, use my trick, and use logic
u will find 1+2+10 = 13 satisfying with the coefficient of x^2 and product is 1*2*10 = 20 satisfying with the constant term.
Now, put x in front of them, factors are:
(x+1)(x+2)(x+12).
U can see one for this that this satisfies one more situation which is
1*2+2*10+10*1 = 32 satisfying coefficient of x.
It will have 3 roots at it is a cubic polynomial,
Now guess those 3 such that their product satisfies the constant term and
their sum satisfies the coefficient of variable with degree 2.
Now, by the method of finding possible roots we can say,
±( 1,2,4,5,10,20) are it's roots.
Now, use my trick, and use logic
u will find 1+2+10 = 13 satisfying with the coefficient of x^2 and product is 1*2*10 = 20 satisfying with the constant term.
Now, put x in front of them, factors are:
(x+1)(x+2)(x+12).
U can see one for this that this satisfies one more situation which is
1*2+2*10+10*1 = 32 satisfying coefficient of x.
it works only when there is cubic polynomial, cant u see
By mistake i wrote 12 instead of 10
If there is like that then we can find new possibilities by b/c concept
dividing the constant term by that coefficient
then the roots will be in fraction and that would be the answer
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