x^3+9x^2+23x+15 using factor theorem factorise
Answers
Answer:
The factor form is x^3+9x^2+23x+15=(x+1)(x+3)(x+5)x
3
+9x
2
+23x+15=(x+1)(x+3)(x+5)
Step-by-step explanation:
Given : Equation x^3+9x^2+23x+15x
3
+9x
2
+23x+15
To find : Factories by factor theorem?
Solution :
Applying rational root theorem state that factor of constant by factor of coefficient of cubic term gives you the possible roots of the equation.
Coefficient of cubic term = 1
Factor = 1
Constant term = 15
Factor of constant term = 1,3,5,15.
Possible roots are \frac{p}{q}= \pm\frac{1,3,5,15}{1}
q
p
=±
1
1,3,5,15
Possible roots are 1,-1,3,-3,5,-5,15,-15.
Substitute all the roots when equation equate to zero then it is the root of the equation.
Put x=-1,
=(-1)^3+9(-1)^2+23(-1)+15=(−1)
3
+9(−1)
2
+23(−1)+15
=-1+9-23+15=−1+9−23+15
=0=0
Put x=-3,
=(-3)^3+9(-3)^2+23(-3)+15=(−3)
3
+9(−3)
2
+23(−3)+15
=-27+81-69+15=−27+81−69+15
=0=0
Put x=-5,
=(-5)^3+9(-5)^2+23(-5)+15=(−5)
3
+9(−5)
2
+23(−5)+15
=-125+225-115+15=−125+225−115+15
=0=0
Therefore, The roots of equation is x=-1,-3,-5.
The factor form is x^3+9x^2+23x+15=(x+1)(x+3)(x+5)x
3
+9x
2
+23x+15=(x+1)(x+3)(x+5)
Answer:
⇒ x³ + 9x + 23x + 15 = (x + 1)(x + 3)(x + 5)
Explanation:
Let the given polynomial,
⇒ p(x) = x³ + 9x² + 23x + 15
Factors of the constant term 15 are ±1, ±3, ±5, and ±15
By factor theorem we find that p(-1) = 0
Then we can say that g(x) = x + 1 is a factor of p(x)
On dividing p(x) by g(x) the quotient will be one of factor of p(x)
So,
x + 1)x³ + 9x² + 23x + 15(x² + 8x + 15)
-
x³ + x²
8x² + 23x
-
8x² + 8x
15x + 15
-
15x + 15
0
We got quotient x² + 8x + 15 which is one of the factor of p(x)
Now, we can say that
⇒ x³ + 9x² + 23x + 15 = (x + 1)(x² + 8x + 15)
Taking L.H.S.
⇒ (x + 1)(x² + 8x + 15)
By middle term splitting,
⇒ (x + 1)[x² + 3x + 5x + 15]
⇒ (x + 1)[x(x + 3) + 5(x + 3)]
⇒ (x + 1)[(x + 3)(x + 5)]
⇒ (x + 1)(x + 3)(x + 5)
Required answer: (x + 1)(x + 3)(x + 5)