p of x=6x-4x^2+3 find x-1
Answers
Answer:
11. Factoring and solving equations
- A. Factor1. Factor 3x2 + 6x if possible.
Look for monomial (single-term) factors first; 3 is a factor of both 3x2
and 6x and so is x . Factor them out to get
3x2 + 6x = 3(x2 + 2x1 = 3x(x+ 2) .
2. Factor x2 + x - 6 if possible.
Here we have no common monomial factors. To get the x2 term
we'll have the form (x +-)(x +-) . Since
(x+A)(x+B) = x2 + (A+B)x + AB ,
we need two numbers A and B whose sum is 1 and whose product is
-6 . Integer possibilities that will give a product of -6 are
-6 and 1, 6 and -1, -3 and 2, 3 and -2.
The only pair whose sum is 1 is (3 and -2) , so the factorization is
x2 + x - 6 = (x+3)(x-2) .
3. Factor 4x2 - 3x - 10 if possible.
Because of the 4x2 term the factored form wli be either
(4x+A)(x +B) or (2x+A)(2x+B) . Because of the -10 the integer possibilities for the pair A, B are
10 and -1 , -10 and 1 , 5 and -2 . -5 and 2 , plus each of
these in reversed order.
Check the various possibilities by trial and error. It may help to write
out the expansions
(4x + A)(x+ B) = 4x2 + (4B+A)x + A8
1 trying to get -3 here
(2x+A)(2x+B) = 4x2 + (2B+ 2A)x + AB
Trial and error gives the factorization 4x2 - 3x - 10 - (4x+5)(x- 2) .
4. Difference of two squares. Since (A + B)(A - B) = - B~ , any
expression of the form A' - B' can be factored. Note that A and B
might be anything at all.
Examples: 9x2 - 16 = (3x1' - 4' = (3x +4)(3x - 4)
x2 - 29 = x2 - (my)* = (x+ JTy)(x- my)
For any of the above examples one could also use the
In the factorization
-
ax2 + bx + c = a(x-Ah-B) ,
the numbers A and B are given by
A,B = -
2a
If the 'discriminant" b2 - 4ac is negative, the polynomial
cannot be factored over the real numbers (e.g. consider x2 + 1).
In Example 2 above, a = 1, b = 1, c = -6 , so
A,B =
2 'l tE3 = '
= 2,-3 , so x +x-6 - (x-2)(~+3). 2 2
5. Factor x3 + 3x2 - 4 if possible.
If plugging x = a into a polynomial yields -zero, then the
polynomial has (x - a) as a factor.
We'll use this fact to try to find factors of x3 + 3x2 - 4 . We look for
factors (x-a) by plugging in various possible a's , choosing those that
are factors of -4 . Try plugging x 1-12 2 4, -4 into
x3+3x2-4. Findthat x-I gives 13+3.12-4 -0. So x-1 isa
factor of x3 + 3x2 - 4 . To factor it out, perform long division:
x2 + 4x + 4 Thus
x-~ lx3 + sx2+ Ox - ~3 + 3x2 - 4 = (x-I)(x~ + 4x + 4).
x3 - .2 But x2 + 4x + 4 can be
4x2 + Ox - 4 factored further as in the
4x2 - 4x examples above;
4x - 4
-
0
we finally get x3 + 3x2 - 4 = (x- l)(x+ 2)(x+2) = (x- 1)(~+2)~ .
s IIA Factor the following polynomials.
1. x2 + 8x + 15 2. 4x2 - 25
3. 49 - 13y - 12 4. x3 + 2x2 - x - 2
5. 4z2 + 42 - 8 6. a2 + 3a + 2
7. Simplify by factoring 3x 2 + 3x - 18
numerator and denominator: 2 4x - 3x - 10
B. Solvina eauatlons
1. Linear or first-degree equations: involving x but not x2 or any
other power of x . Collect x-terms on one side, constant terms on the
other.
ExamDle x+3=7x-4
x + (-7x1 = -4 + (-3)
-6x = -7
x = 7/6
2. Quadratic equations: involving x2 but no higher power of x . These are solved by factoring and/or use of the quadratic formula:
The equation ax2 + bx + c = 0 (a - 0)
has solutions x =
2a
If b2 - 4ac is negative, the equation has no real solutions.
Exam& Solve x2-2x-3=0 for x.
Method: Factoring. x2 - 2x - 3 = (x- 3)(x+1) = 0 . Since a product of two numbers is zero if and only if one of the two
numbers is zero, we must have
x - 3 = 0 or x + I = 0 . So the solutions are x = 3. -1 .
Mefhod: Quadratic formula. a = 1 , b = -2 . c = -3 .
2
X = -(-2) k \/(-2) - 4(1)(-3) . 2 k &i - 2 k 4
2
= 3 or -1
2(1) 2
3. Other types of equations.
14 (a) Solve - - - - I
x+2 x-4
Multiply both sides by common denominator (x+ 2)(x - 4) to get
14(x-4) - 1(x+2) - (x+2)(x-4).
Expand and simplify. Get a quadratic equation so put all terms on one
side. 14x-56-x-2= x2-2x-8
x2 - 15x + 50 = 0
Now factor (or use quadratic formula).
(x-10)(x-5) = 0, x-los 0 or x-5= 0, x= 10 or 5.
(b) Solve x3 - 2x2 - 5x + 6 = 0 .
The idea is much the same as in Example 5 of part A where we used
the fact about factoring polynomials. Try x = 1, -1,2, -2,3, -3,6, -6 . As soon as one of these possibilities satisfies the equation we have a factor.
It happens that x = 1 is a solution. By long division we get:
x3 - 2x2 - 5x + 6 = (x-1)(x2 - x - 6) = (x-I)(x-3)(~+2) = 0 ,
so x = 1,3,or -2.
(c) Solve Jx+T = x .
Start by squaring both sides, but this may lead to extraneous roots so
we'll have to check answers at the end.
x2 - x- 2 = (~-2)(x+1) = 0. so x = 2or-1.
Check in original equation: fi+y - 2, OK; ./-- - I, not -1 , so
reject x = -1 ; only solution is x = 2 .
s IIB Solve the following equations.
1 2 (solve for g in terms 2. s = 7 gt of s and t .)
3. s = xgt2 (solvefor t interrnsof s,g)
4. x2 - (x-2)2x = 4 5. x=-4,+3 = 0
1. Linear systems of equations.
ExamDlc Find all values of x and y that satisfy
Step-by-step explanation: