the sum of the square of n numbers is 2n^3 + 3n^2+n /.6 .factorise this expression
CLASS 8 MATHS
Answers
1.1 Factoring: n2-3
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : 3 is not a square !!
Ruling : Binomial can not be factored as the difference of two perfect squares.
Equation at the end of step
1
:
(((5•(n2))•(n-6))-(2n•(((3•(n2))+n)-6)))+7•(n2-3)
STEP
2
:
Equation at the end of step
2
:
(((5•(n2))•(n-6))-(2n•((3n2+n)-6)))+7•(n2-3)
STEP
3
:
Trying to factor by splitting the middle term
3.1 Factoring 3n2+n-6
The first term is, 3n2 its coefficient is 3 .
The middle term is, +n its coefficient is 1 .
The last term, "the constant", is -6
Step-1 : Multiply the coefficient of the first term by the constant 3 • -6 = -18
Step-2 : Find two factors of -18 whose sum equals the coefficient of the middle term, which is 1 .
-18 + 1 = -17
-9 + 2 = -7
-6 + 3 = -3
-3 + 6 = 3
-2 + 9 = 7
-1 + 18 = 17
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step
3
:
(((5•(n2))•(n-6))-2n•(3n2+n-6))+7•(n2-3)
STEP
4
:
Equation at the end of step
4
:
((5n2•(n-6))-2n•(3n2+n-6))+7•(n2-3)
STEP
5
:
Equation at the end of step 5
(5n2•(n-6)-2n•(3n2+n-6))+7•(n2-3)
STEP
6
:
STEP
7
:
Pulling out like terms
7.1 Pull out like factors :
-n3 - 25n2 + 12n - 21 =
-1 • (n3 + 25n2 - 12n + 21)
Checking for a perfect cube :
7.2 n3 + 25n2 - 12n + 21 is not a perfect cube
Trying to factor by pulling out :
7.3 Factoring: n3 + 25n2 - 12n + 21
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: -12n + 21
Group 2: n3 + 25n2
Pull out from each group separately :
Group 1: (4n - 7) • (-3)
Group 2: (n + 25) • (n2)
Bad news !! Factoring by pulling out fails :
The groups have no common factor and can not be added up to form a multiplication.
Polynomial Roots Calculator :
7.4 Find roots (zeroes) of : F(n) = n3 + 25n2 - 12n + 21
Polynomial Roots Calculator is a set of methods aimed at finding values of n for which F(n)=0
Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers n which can be expressed as the quotient of two integers
The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient
In this case, the Leading Coefficient is 1 and the Trailing Constant is 21.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,3 ,7 ,21
Let us test ....
P Q P/Q F(P/Q) Divisor
-1 1 -1.00 57.00
-3 1 -3.00 255.00
-7 1 -7.00 987.00
-21 1 -21.00 2037.00
1 1 1.00 35.00
3 1 3.00 237.00
7 1 7.00 1505.00
21 1 21.00 20055.00
Polynomial Roots Calculator found no rational roots
Final result :
-n3 - 25n2 + 12n - 21
Answer:
STEP1:Equation at the end of step 1
(((2•(n3))-3n2)+n) (——————————————————-n)+2 6
STEP 2 :
Equation at the end of step2:
((2n3 - 3n2) + n) (————————————————— - n) + 2 6
STEP3:
2n3 - 3n2 + n Simplify ————————————— 6
STEP4:Pulling out like terms
4.1 Pull out like factors :
2n3 - 3n2 + n = n • (2n2 - 3n + 1)
Trying to factor by splitting the middle term
4.2 Factoring 2n2 - 3n + 1
The first term is, 2n2 its coefficient is 2 .
The middle term is, -3n its coefficient is -3 .
The last term, "the constant", is +1
Step-1 : Multiply the coefficient of the first term by the constant 2 • 1 = 2
Step-2 : Find two factors of 2 whose sum equals the coefficient of the middle term, which is -3 .
-2 + -1 = -3 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -2 and -1
2n2 - 2n - 1n - 1
Step-4 : Add up the first 2 terms, pulling out like factors :
2n • (n-1)
Add up the last 2 terms, pulling out common factors :
1 • (n-1)
Step-5 : Add up the four terms of step 4 :
(2n-1) • (n-1)