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## 12y^4+33y^3+6y

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# 1 Answer

Hi Tom,

12y^4+33y^3+6y

=12y(y^3+11/4* y^2+1/2)

Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integers.

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integers.

Root test:

If a polynomial zeroes for a rational number p/q then p is a factor of the trailing constant & q is a factor of the leading coeff

for f(y)= y^3+11/4* y^2+1/2

Leading coeff is 1 , the trailing constant is 1/2

factors of 1 is 1

factors of 1/2 is 1,1/2.

p q p/q f(p/q)

-1 1 -1 (-1)^3+11/4*(-1)^2+1/2= -1+11/4+1/2=9/4

-1/2 1 -1/2 (-1/2)^3+11/4* (-1/2)^2+1/2 = -1/8+11/16+1/2= 17/16

Polynomial Roots Calculator found no rational roots.

Let y+1/4 is one of the factor,then

Divide y^3+11/4*y^2+42/64-10/64 by y+1/4

=(y+1/4)(y^2+5y/2-5/8)+21/32

=(y+1/4)(8y^2+20y-5 /8)+21/32

= 1/8(y+1/4)(8y^2+20y-5)+21/32

using quad formula, -20+sq root of 400+ 160 /16= -20+sq root of 560 /16=-20+4 sq root of 35 /16=4(-5+sq root of 35)/16=-5 +sq root of 35 /4=-5/4 +sq root of 35/4

or -20-sq root of 400+160 /16= -5-sq root of 35 /4=-5/4 -sq root of 35/4

8y^2+20y-5=(y+5/4-sq root of 35/4)(y+5/4+sq root of 35/4)

12y^4+33y^3+6y

=12y [(y+1/4)(8y^2+20y-5)/8+21/32]

=12y [(y+1/4)(y+5/4-sq root of 35/4)(y+5/4+sq root of 35/4)+21/32]

=12y(y+1/4)(y^2+5y/4+sq root of 35y/4+5y/4+25/16+sq root of 35*5/4-sq root of 35*y/4-5/4*sq root of 35/4-35/16)+21/32]

=12y[(y+1/4)(y^2+5y/2-5/8)+21/32]

Meena from Strongsville,Oh