Factorise the polynomial X^3+13X^2+32X-36 if one of it's zero is -9
and first prove that -9 is the zero of polynomial.
Answers
P(x)=x³+13x²+32x-36
one of its zero is -9
zero of polynomial written as x+9
Factorise the given polynomial
Proved that -9 is zero of this polynomial
Given , the zero of polynomial we put the in the given polynomial to prove that -9 is the zero polynomial
=>P(x)=x³+13x²+32x-36
=>P(-9)=(-9)³+13×(-9)²+32×(-9)-36
=>P(-9)=-729+13×81-32×9-36
=>P(-9)=1053-729-288-36
=>P(-9)=1053-1053
=>P(-9)=0
Now we have to divide P(x) by x+9
x+9)x³+13x²+32x-36(x²+4x-4
x³+9x²
- -
4x²+32x-36
4x²+36x
- +
-4x-36
-4x-36
+ +
× ×
Now, we have to factorise here with help of divisor and quotient
=>(x+9)(x²+4x-4)
By quardric equation to solve
=>X²+4x-4
By quardric equation
a=1 b=4 c=-4
D=b²-4ac
D=4²-4*1*(-4)
D=16+16=32
X=-b+-√D/2
X=-4-√32/2=>4-4√2/2=>4(1-√2)/2=>2-2√2 X=2+2√2
=>(x+9)(x+2-2√2)(x+2+2√2)
Hence,
The required answer is (x+9)(x+2-2√2)(x-2+2√2)
Answer:
GivenPolynomial:
P(x)=x³+13x²+32x-36
one of its zero is -9
zero of polynomial written as x+9
\large{\underline{\red{\rm{To\: Find:}}}}
ToFind:
Factorise the given polynomial
Proved that -9 is zero of this polynomial
\large{\underline{\orange{\rm{Solution:}}}}
Solution:
Given , the zero of polynomial we put the in the given polynomial to prove that -9 is the zero polynomial
=>P(x)=x³+13x²+32x-36
=>P(-9)=(-9)³+13×(-9)²+32×(-9)-36
=>P(-9)=-729+13×81-32×9-36
=>P(-9)=1053-729-288-36
=>P(-9)=1053-1053
=>P(-9)=0
Now we have to divide P(x) by x+9
x+9)x³+13x²+32x-36(x²+4x-4
x³+9x²
- -
4x²+32x-36
4x²+36x
- +
-4x-36
-4x-36
+ +
× ×
Now, we have to factorise here with help of divisor and quotient
=>(x+9)(x²+4x-4)
By quardric equation to solve
=>X²+4x-4
By quardric equation
a=1 b=4 c=-4
D=b²-4ac
D=4²-4*1*(-4)
D=16+16=32
X=-b+-√D/2
X=-4-√32/2=>4-4√2/2=>4(1-√2)/2=>2-2√2 X=2+2√2
=>(x+9)(x+2-2√2)(x+2+2√2)
Hence,
The required answer is (x+9)(x+2-2√2)(x-2+2√2)