Find the co-efficient of x^2: x^3+13x^2+50x+56
Answers
Answer:
coefficient of x²= 13
Step-by-step explanation:
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We think you wrote:
x^3-13x^2+50x-56=0
3 result(s) found
x= 7
x= 4
x= 2
See steps
Step by Step Solution:
Step by step solution :
STEP 1 : Equation at the end of step 1
(((x3) - 13x2) + 50x) - 56 = 0
STEP 2 : Checking for a perfect cube
2.1 x3-13x2+50x-56 is not a perfect cube
Trying to factor by pulling out :
2.2 Factoring: x3-13x2+50x-56
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 50x-56
Group 2: x3-13x2
Pull out from each group separately :
Group 1: (25x-28) • (2)
Group 2: (x-13) • (x2)
Polynomial Roots Calculator :
2.3 Find roots (zeroes) of : F(x) = x3-13x2+50x-56
Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0
In this case, the Leading Coefficient is 1 and the Trailing Constant is -56.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4 ,7 ,8 ,14 ,28 ,56
Let us test ....
P Q P/Q F(P/Q) Divisor
-1 1 -1.00 -120.00
-2 1 -2.00 -216.00
-4 1 -4.00 -528.00
-7 1 -7.00 -1386.00
-8 1 -8.00 -1800.00
-14 1 -14.00 -6048.00
-28 1 -28.00 -33600.00
-56 1 -56.00 -219240.00
1 1 1.00 -18.00
2 1 2.00 0.00 x-2
4 1 4.00 0.00 x-4
7 1 7.00 0.00 x-7
8 1 8.00 24.00
14 1 14.00 840.00
28 1 28.00 13104.00
56 1 56.00 137592.00
The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p
In our case this means that
x3-13x2+50x-56
can be divided by 3 different polynomials,including by x-7
Polynomial Long Division :
2.4 Polynomial Long Division
Dividing : x3-13x2+50x-56
("Dividend")
By : x-7 ("Divisor")
dividend x3 - 13x2 + 50x - 56
- divisor * x2 x3 - 7x2
remainder - 6x2 + 50x - 56
- divisor * -6x1 - 6x2 + 42x
remainder 8x - 56
- divisor * 8x0 8x - 56
remainder 0
Quotient : x2-6x+8 Remainder: 0
Trying to factor by splitting the middle term
2.5 Factoring x2-6x+8
The first term is, x2 its coefficient is 1 .
The middle term is, -6x its coefficient is -6 .
The last term, "the constant", is +8
Step-1 : Multiply the coefficient of the first term by the constant 1 • 8 = 8
Step-2 : Find two factors of 8 whose sum equals the coefficient of the middle term, which is -6 .
-8 + -1 = -9
-4 + -2 = -6 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -4 and -2
x2 - 4x - 2x - 8
Step-4 : Add up the first 2 terms, pulling out like factors :
x • (x-4)
Add up the last 2 terms, pulling out common factors :
2 • (x-4)
Step-5 : Add up the four terms of step 4 :
(x-2) • (x-4)
Which is the desired factorization
Equation at the end of step 2 :
(x - 2) • (x - 4) • (x - 7) = 0
STEP 3 : Theory - Roots of a product
3.1 A product of several terms equals zero.
We shall now solve each term = 0 separately
Solving a Single Variable Equation:
3.2 Solve : x-2 = 0
Add 2 to both sides of the equation :
x = 2
Solving a Single Variable Equation:
3.3 Solve : x-4 = 0
Add 4 to both sides of the equation :
x = 4
Solving a Single Variable Equation:
3.4 Solve : x-7 = 0
Add 7 to both sides of the equation :
x = 7
Supplement : Solving Quadratic Equation Directly
Solving x2-6x+8 = 0 directly
Parabola, Finding the Vertex:
4.1 Find the Vertex of y = x2-6x+8
For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 3.0000
y = 1.0 * 3.00 * 3.00 - 6.0 * 3.00 + 8.0
or y = -1.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = x2-6x+8
Axis of Symmetry (dashed) {x}={ 3.00}
Vertex at {x,y} = { 3.00,-1.00}
x -Intercepts (Roots) :
Root 1 at {x,y} = { 2.00, 0.00}
Root 2 at {x,y} = { 4.00, 0.00}
4.2 Solving x2-6x+8 = 0 by Completing The Square .
Subtract 8 from both side of the equation :
x2-6x = -8
Add 9 to both sides of the equation :
On the right hand side we have :
-8 + 9 or, (-8/1)+(9/1)
The common denominator of the two fractions is 1 Adding (-8/1)+(9/1) gives 1/1
So adding to both sides we finally get :
x2-6x+9 = 1
Adding 9 has completed the left hand side into a perfect square :
x2-6x+9 =
(x-3) • (x-3) =
(x-3)2
x2-6x+9 = 1 and
x2-6x+9 = (x-3)2
then, according to the law of transitivity,
(x-3)2 = 1
Note that the square root of
(x-3)2 is
(x-3)2/2 =
(x-3)1 =
x-3
Now, applying the Square Root Principle to we get:
x-3 = √ 1
Add 3 to both sides to obtain:
x = 3 + √ 1
Since a square root has two values, one positive and the other negative
x2 - 6x + 8 = 0
has two solutions:
x = 3 + √ 1
or
x = 3 - √ 1
Solve Quadratic Equation using the Quadratic Formula
4.3 Solving x2-6x+8 = 0 by the Quadratic Formula .
According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , is given by :
- B ± √ B2-4AC
x = ————————
2A
In our case, A = 1
B = -6
C = 8
Accordingly, B2 - 4AC =
36 - 32 =
4
Applying the quadratic formula :
6 ± √ 4
x = ————
2
Can √ 4 be simplified ?
Yes! The prime factorization of 4 is
2•2
√ 4 = √ 2•2 =
± 2 • √ 1 =
± 2
So now we are looking at:
x = ( 6 ± 2) / 2
Two real solutions:
x =(6+√4)/2=3+= 4.000
or:
x =(6-√4)/2=3-= 2.000
Three solutions were found :
x = 7
x = 4
x = 2