Find the zeros ofcthe polynomials f(x)=x3−12x2+39x−28, if it's is given that the zeros in AP

f(x)=x3−12x2+39x−28 Let x,y,z are zeros of given polynomial. we know that, if x,y,z, are zero of cubic polynomial ax3+bx2+cx+d then, x+y+z=−ba xy+yz+zx=ca xyz=−da now given polynomial is, x3−12x2+39x−28 Here, a=1,b=−12,c=39,d=−28 x+y+z=−ba=−121=12 xy+yz+zx=ca=391=39 xyz=−da=−−281=28 Now, given that polynomial zeros are in A.P., x=a−d.y=a.z=a+d put these value of x,y,z in equation, we get (a−d)+a+(a+d)=12⇒3a=12⇒a=4 put value of x,y,z in equation we get (a−d)+a+(a+d)=28⇒a(a2−d2)=28 ⇒a(a2−d2)=28 now a=4, 4(16−d2)=28 d2=9⇒d=±3 we know, x=a−d=4−5=1 y=a=4 z=a+d=4+3=7 The polynomial zero are, 1,4,7

Find out the answer to "Find the zeros ofcthe polynomials f(x)=x3−12x2+39x−28, if it's..."

CoormaBak9

Answered question

2021-09-19

Find the zeros ofcthe polynomials

f (x) = x^{3} - 12 x^{2} + 39 x - 28

, if it's is given that the zeros in AP

Answer & Explanation

coffentw

Skilled2021-09-20Added 103 answers

f (x) = x^{3} - 12 x^{2} + 39 x - 28

Let x,y,z are zeros of given polynomial.
we know that,
if x,y,z, are zero of cubic polynomial

a x^{3} + b x^{2} + c x + d

then,

x + y + z = \frac{- b}{a}

x y + y z + z x = \frac{c}{a}

x y z = \frac{- d}{a}

now given polynomial is,

x^{3} - 12 x^{2} + 39 x - 28

Here,

a = 1, b = - 12, c = 39, d = - 28

x + y + z = \frac{- b}{a} = \frac{- 12}{1} = 12

x y + y z + z x = \frac{c}{a} = \frac{39}{1} = 39

x y z = - \frac{d}{a} = - \frac{- 28}{1} = 28

Now, given that polynomial zeros are in A.P.,

x = a - d . y = a . z = a + d

put these value of x,y,z in equation, we get

(a - d) + a + (a + d) = 12 \Rightarrow 3 a = 12 \Rightarrow a = 4

put value of x,y,z in equation we get

(a - d) + a + (a + d) = 28 \Rightarrow a (a^{2} - d^{2}) = 28

\Rightarrow a (a^{2} - d^{2}) = 28

now a=4,

4 (16 - d^{2}) = 28

d^{2} = 9 \Rightarrow d = \pm 3

we know,

x = a - d = 4 - 5 = 1

y = a = 4

z = a + d = 4 + 3 = 7

The polynomial zero are, 1,4,7

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Precalculus

Didn't find what you were looking for?