Online calculator.
Isolating the square of a binomial and factoring a square trinomial.
This math program distinguishes the square binomial from the square trinomial, i.e. does a transformation like: \(ax^2+bx+c \rightarrow a(x+p)^2+q \) and factorizes quadratic trinomial : \(ax^2+bx+c \rightarrow a(x+n)(x+m) \)
Those. the problems boil down to finding the numbers \(p, q\) and \(n, m\)
The program not only gives the answer to the problem, but also displays the solution process.
This program may be useful for high school students secondary schools in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as quickly as possible? homework in mathematics or algebra? In this case, you can also use our programs with detailed solutions.
This way you can conduct your own training and/or training of yours. younger brothers or sisters, while the level of education in the field of problems being solved increases.
If you are not familiar with the rules for entering a quadratic trinomial, we recommend that you familiarize yourself with them.
Rules for entering a quadratic polynomial
Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q\), etc.
Numbers can be entered as whole or fractional numbers.
Moreover, fractional numbers can be entered not only as a decimal, but also as an ordinary fraction.
Rules for entering decimal fractions.
In decimals fraction can be separated from the whole by either a period or a comma.
For example, you can enter decimals like this: 2.5x - 3.5x^2
Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering numerical fraction The numerator is separated from the denominator by a division sign: /
The whole part is separated from the fraction by the ampersand sign: &
Input: 3&1/3 - 5&6/5x +1/7x^2
Result: \(3\frac(1)(3) - 5\frac(6)(5) x + \frac(1)(7)x^2\)
When entering an expression you can use parentheses. In this case, when solving, the introduced expression is first simplified.
For example: 1/2(x-1)(x+1)-(5x-10&1/2)
Example detailed solution
Isolating the square of a binomial.$$ ax^2+bx+c \rightarrow a(x+p)^2+q $$ $$2x^2+2x-4 = $$ $$2x^2 +2 \cdot 2 \cdot\left( \frac(1)(2) \right)\cdot x+2 \cdot \left(\frac(1)(2) \right)^2-\frac(9)(2) = $$ $$2\left (x^2 + 2 \cdot\left(\frac(1)(2) \right)\cdot x + \left(\frac(1)(2) \right)^2 \right)-\frac(9 )(2) = $$ $$2\left(x+\frac(1)(2) \right)^2-\frac(9)(2) $$ Answer:$$2x^2+2x-4 = 2\left(x+\frac(1)(2) \right)^2-\frac(9)(2) $$ Factorization.$$ ax^2+bx+c \rightarrow a(x+n)(x+m) $$ $$2x^2+2x-4 = $$
$$ 2\left(x^2+x-2 \right) = $$
$$ 2 \left(x^2+2x-1x-1 \cdot 2 \right) = $$ $$ 2 \left(x \left(x +2 \right) -1 \left(x +2 \right ) \right) = $$ $$ 2 \left(x -1 \right) \left(x +2 \right) $$ Answer:$$2x^2+2x-4 = 2 \left(x -1 \right) \left(x +2 \right) $$
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A little theory.
Isolating the square of a binomial from a square trinomial
If the square trinomial ax 2 +bx+c is represented as a(x+p) 2 +q, where p and q are real numbers, then they say that from square trinomial, the square of the binomial is highlighted.
From the trinomial 2x 2 +12x+14 we extract the square of the binomial.
\(2x^2+12x+14 = 2(x^2+6x+7) \)
To do this, imagine 6x as a product of 2*3*x, and then add and subtract 3 2. We get:
$$ 2(x^2+2 \cdot 3 \cdot x + 3^2-3^2+7) = 2((x+3)^2-3^2+7) = $$ $$ = 2 ((x+3)^2-2) = 2(x+3)^2-4 $$
That. We extract the square binomial from the square trinomial, and showed that:
$$ 2x^2+12x+14 = 2(x+3)^2-4 $$
Factoring a quadratic trinomial
If the square trinomial ax 2 +bx+c is represented in the form a(x+n)(x+m), where n and m are real numbers, then the operation is said to have been performed factorization of a quadratic trinomial.
Let us show with an example how this transformation is done.
Let's factor the quadratic trinomial 2x 2 +4x-6.
Let us take the coefficient a out of brackets, i.e. 2:
\(2x^2+4x-6 = 2(x^2+2x-3) \)
Let's transform the expression in brackets.
To do this, imagine 2x as the difference 3x-1x, and -3 as -1*3. We get:
$$ = 2(x^2+3 \cdot x -1 \cdot x -1 \cdot 3) = 2(x(x+3)-1 \cdot (x+3)) = $$
$$ = 2(x-1)(x+3) $$
That. We factored the quadratic trinomial, and showed that:
$$ 2x^2+4x-6 = 2(x-1)(x+3) $$
Note that factoring a quadratic trinomial is possible only if the quadratic equation corresponding to this trinomial has roots.
Those. in our case, it is possible to factor the trinomial 2x 2 +4x-6 if the quadratic equation 2x 2 +4x-6 =0 has roots. In the process of factorization, we established that the equation 2x 2 + 4x-6 = 0 has two roots 1 and -3, because with these values, the equation 2(x-1)(x+3)=0 turns into a true equality.
Definition
Expressions of the form 2 x 2 + 3 x + 5 are called quadratic trinomials. IN general case a square trinomial is an expression of the form a x 2 + b x + c, where a, b, c a, b, c - arbitrary numbers, and a ≠ 0.
Consider the quadratic trinomial x 2 - 4 x + 5. Let's write it in this form: x 2 - 2 · 2 · x + 5. Let's add 2 2 to this expression and subtract 2 2, we get: x 2 - 2 · 2 · x + 2 2 - 2 2 + 5. Note that x 2 - 2 2 x + 2 2 = (x - 2) 2, so x 2 - 4 x + 5 = (x - 2) 2 - 4 + 5 = (x - 2) 2 + 1 . The transformation we made is called “isolating a perfect square from a quadratic trinomial”.
Highlight perfect square from the quadratic trinomial 9 x 2 + 3 x + 1.
Note that 9 x 2 = (3 x) 2 , `3x=2*1/2*3x`. Then `9x^2+3x+1=(3x)^2+2*1/2*3x+1`. Add and subtract `(1/2)^2` to the resulting expression, we get
`((3x)^2+2*1/2*3x+(1/2)^2)+1-(1/2)^2=(3x+1/2)^2+3/4`.
We will show how the method of isolating a perfect square from a quadratic trinomial is used to factorize a square trinomial.
Factor the quadratic trinomial 4 x 2 - 12 x + 5.
We select the perfect square from the quadratic trinomial: 2 x 2 - 2 · 2 x · 3 + 3 2 - 3 2 + 5 = 2 x - 3 2 - 4 = (2 x - 3) 2 - 2 2. Now we apply the formula a 2 - b 2 = (a - b) (a + b) , we get: (2 x - 3 - 2) (2 x - 3 + 2) = (2 x - 5) (2 x - 1 ) .
Factor the quadratic trinomial - 9 x 2 + 12 x + 5.
9 x 2 + 12 x + 5 = - 9 x 2 - 12 x + 5. Now we notice that 9 x 2 = 3 x 2, - 12 x = - 2 3 x 2.
We add the term 2 2 to the expression 9 x 2 - 12 x, we get:
3 x 2 - 2 3 x 2 + 2 2 - 2 2 + 5 = - 3 x - 2 2 - 4 + 5 = 3 x - 2 2 + 4 + 5 = - 3 x - 2 2 + 9 = 3 2 - 3 x - 2 2 .
We apply the formula for the difference of squares, we have:
9 x 2 + 12 x + 5 = 3 - 3 x - 2 3 + (3 x - 2) = (5 - 3 x) (3 x + 1) .
Factor the quadratic trinomial 3 x 2 - 14 x - 5 .
We cannot represent the expression 3 x 2 as the square of some expression, because we have not yet studied this in school. You will go through this later, and in Task No. 4 we will study square roots. Let's show how you can factor a given quadratic trinomial:
`3x^2-14x-5=3(x^2-14/3 x-5/3)=3(x^2-2*7/3 x+(7/3)^2-(7/3) ^2-5/3)=`
`=3((x-7/3)^2-49/9-5/3)=3((x-7/3)^2-64/9)=3((x-7/3)^ 2-8/3)^2)=`
`=3(x-7/3-8/3)(x-7/3+8/3)=3(x-5)(x+1/3)=(x-5)(3x+1) `.
We'll show you how to use the perfect square method to find the largest or smallest value of a quadratic trinomial.
Consider the quadratic trinomial x 2 - x + 3. Select a complete square:
`(x)^2-2*x*1/2+(1/2)^2-(1/2)^2+3=(x-1/2)^2+11/4`. Note that when `x=1/2` the value of the quadratic trinomial is `11/4`, and when `x!=1/2` the value of `11/4` is added positive number, so we get a number greater than `11/4`. Thus, smallest value quadratic trinomial is `11/4` and it is obtained when `x=1/2`.
Find the largest value of the quadratic trinomial - 16 2 + 8 x + 6.
We select a perfect square from a quadratic trinomial: - 16 x 2 + 8 x + 6 = - 4 x 2 - 2 4 x 1 + 1 - 1 + 6 = - 4 x - 1 2 - 1 + 6 = - 4 x - 1 2 + 7 .
When `x=1/4` the value of the quadratic trinomial is 7, and when `x!=1/4` a positive number is subtracted from the number 7, that is, we get a number less than 7. So the number 7 is highest value quadratic trinomial, and it is obtained when `x=1/4`.
Factor the numerator and denominator of the fraction `(x^2+2x-15)/(x^2-6x+9)` and reduce the fraction.
Note that the denominator of the fraction x 2 - 6 x + 9 = x - 3 2. Let's factorize the numerator of the fraction using the method of isolating a complete square from a square trinomial. x 2 + 2 x - 15 = x 2 + 2 x 1 + 1 - 1 - 15 = x + 1 2 - 16 = x + 1 2 - 4 2 = = (x + 1 + 4) (x + 1 - 4) = (x + 5) (x - 3) .
This fraction led to the form `((x+5)(x-3))/(x-3)^2` after reduction by (x - 3) we get `(x+5)/(x-3)`.
Factor the polynomial x 4 - 13 x 2 + 36.
Let us apply the method of isolating a complete square to this polynomial. `x^4-13x^2+36=(x^2)^2-2*x^2*13/2+(13/2)^2-(13/2)^2+36=(x^ 2-13/2)^2-169/4+36=(x^2-13/2)^2-25/4=`
x called
1.2.3. Using abbreviated multiplication identities
Example. Factor x 4 16.
x 4 16x 2 2 42 x 2 4x 2 4x 2x 2x 2 4 .
1.2.4. Factoring a polynomial using its roots
Theorem. Let the polynomial P x have root x 1 . Then this polynomial can be factorized as follows: P x x x 1 S x , where S x is some polynomial whose degree is one less
values alternately into the expression for P x. We obtain that when x 2 you-
the expression will turn to 0, that is, P 2 0, which means x 2 is the root of a multi-
member. Divide the polynomial P x by x 2 .
X 3 3x 2 10x 24 | ||
x 32 x 2 | 24 10 x | x2 x12 |
12x 2412x 24
P x x 2 x2 x12 x2 x2 3 x4 x12 x2 x x3 4 x3
x2 x3 x4
1.3. Selecting a complete square
The method for selecting a complete square is based on the use of the formulas: a 2 2ab b 2 a b 2 ,a 2 2ab b 2 a b 2 .
Isolating a complete square is an identity transformation in which a given trinomial is represented as a b 2 the sum or difference of the square of the binomial and some numerical or alphabetic expression.
Square trinomial relative to variable size there is an expression of the form
ax 2 bx c , where a ,b and c – given numbers, and a 0 . | |||||||||||||
Let us transform the quadratic trinomial ax 2 bx c as follows. | x2: |
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coefficient | |||||||||||||
Then we represent the expression b x as 2b x (twice the product
x ):a x | ||||||||||||||||
To the expression in parentheses we add and subtract the number from it
which is the square of a number | As a result we get: | |||||||||||||||||||||||||||||||||||||||||
Noticing now that | We get | |||||||||||||||||||||||||||||||||||||||||
4a 2 | ||||||||||||||||||||||||||||||||||||||||||
Example. Select a complete square. | 2 x 12 | |||||||||||||||||||||||||||||||||||||||||
2x 2 4x 5 2x 2 2x 5 | 2 x 2 2x 1 15 | |||||||||||||||||||||||||||||||||||||||||
2 x 12 7.
4 a 2,
1.4. Polynomials in several variables
Polynomials in several variables, like polynomials in one variable, can be added, multiplied and raised to a natural power.
Important identical transformation a polynomial in several variables is factorization. Here, such methods of factorization as removal are used common multiplier beyond brackets, grouping, using abbreviated multiplication identities, highlighting the complete square, introducing auxiliary variables.
1. Factor the polynomial P x ,y 2x 5 128x 2 y 3 .
2 x 5128 x 2y 32 x 2x 364 y 32 x 2x 4 y x 24 xy 16 y 2.
2. Factor P x ,y ,z 20x 2 3yz 15xy 4xz . Let's apply the grouping method
20 x2 3 yz15 xy4 xz20 x2 15 xy4 xz3 yz5 x4 x3 y z4 x3 y
4 x3 y5 x z.
3. Factor P x ,y x 4 4y 4 . Let's select a complete square:
x 4y 4x 44 x 2y 24 y 24 x 2y 2x 22 y 2 2 4 x 2y 2
x2 2 y2 2 xy x2 2 y2 2 xy.
1.5. Properties of a degree with any rational exponent
Degree with any rational indicator has the following properties:
1. a r 1a r 2a r 1r 2,
a r 1a r 2a r 1r 2, |
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3. a r 1r 2 a r 1r 2, |
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4. abr 1 ar 1 br 1, |
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a r 1 | ar 1 |
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br 1 |
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where a 0;b 0;r 1;r 2 are arbitrary rational numbers.
1. Multiply 8 | x 3 12x 7. | |||||||||||||||||||||||||||||||||||||||||||||||||||||
24 x 23. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||
8 x 3 12 x 7 x 8x 12x 8 12x 24 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||
2. Factorize | a 2x 3 | |||||||||||||||||||||||||||||||||||||||||||||||||||||
1.6. Exercises to do on your own
1. Perform actions using abbreviated multiplication formulas. 1) a 52 ;
2) 3 a 72 ;
3) a nb n2 .
4) 1 x 3 ;
3 y 3 ; | |||||
7) 8 a 2 8a 2 ;
8) a nb ka kb na nb ka kb n.
9) a 2 b a2 2 ab4 b2 ;
10) a 3a 2 3a 9 ;
11) a 2b 2a 4a 2b 2b 4. 3
2. Calculate using abbreviated multiplication identities:
1) 53 2 432 ;
2) 22,4 2 22,32 ;
4) 30 2 2 ;
5) 51 2 ;
6) 99 2 ;
7) 17 2 2 17 23 232 ;
8) 85 2 2 85 15 152 .
3. Prove the identities:
1). x 2 13 3x 2 x 12 6x x 1 11x 3 32 2;
2) a 2b 2 2 2 ab 2 a 2b 2 2 ;
3) a 2 b2 x2 y2 ax by2 bx ay2 .
4. Factor the following polynomials:
1) 3 x a2 a2;
2) ac 7 bc3 a21 b;
3) 63 m 4n 327 m 3n 445 m 5n 7;
4) 5 b2 c3 2 bc2 k2 k2 ;
5) 2 x3 y2 3 yz2 2 x2 yz3 z3 ;
6) 24 ax38 bx12 a19 b;
7) 25 a 21 b 2q 2;
8) 9 5 a 4b 2 64a 2 ;
9) 121 n 2 3n 2t 2 ;
10) 4 t 2 20tn 25n 2 36;
11) p 4 6 p2 k9 k2 ;
12) 16 p 3 q 8 72p 4 q 7 81p 5 q 6 ;
13) 6 x 3 36x 2 72x 48;
14) 15 ax 3 45 ax 2 45 ax 15 a ;
15) 9 a 3 n 1 4.5a 2 n 1 ;
16) 5 p 2 n q n 15p 5 n q 2 n ;
17) 4 a 7b 232 a 4b 5;
18) 7 x 24 y 2 2 3 x 28 y 2 2 ;
19) 1000 t 3 27t 6 .
5. Calculate in the simplest way:
1) 59 3 413 ;
2) 67 3 523 67 52. 119
6. Find the quotient and remainder of a polynomial P x by polynomialQ x: 1)P x 2x 4 x 3 5;Q x x 3 9x ;
2) P x 2 x 2; Q x x3 2 x2 x; 3) P x x6 1; Q x x4 4 x2 .
7. Prove that the polynomial x 2 2x 2 has no real roots.
8. Find the roots of the polynomial:
1) x 3 4 x;
2) x 3 3x 2 5x 15.
9. Factor:
1) 6 a 2 a 5 5a 3 ;
2) x 2 x 3 2x 32 4x 3 3x 2 ;
3) x 3 6x 2 11x 6.
10. Solve equations by isolating a complete square:
1) x 2 2x 3 0;
2) x 2 13x 30 0 .
11. Find the meanings of expressions:
4 3 85 | ||||
16 6 | ||||
2 520 9 519 | ||||
1254
3) 5 3 25 7 ;
4) 0,01 2 ;
5) 06 .
12. Calculate:
16 0,25 | 16 0,25 | |||||||||||||||||||||||
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