Question

Помогите пожалуйста. Представьте данный многочлен в виде квадрата двухчлена: а)m^2+2mn+n^2 б)4a^2-4ab+b^2 в)100 p^2-60p+9 г)49x^2 y^2-14xy+1 д)t^4-26t^2+169 (^ это степень если что) )

Answer 1

а)m^2+2mn+n^2=(m+n)²
б)4a^2-4ab+b^2=(2a-b)²
в)100 p^2-60p+9=(10p-3)²
г)49x^2 y^2-14xy+1=(7x-1)²
д)t^4-26t^2+169=(t²-13)²

Answer 2

А) m² + 2mn+ n²= (m+ n)²
б) 4a² -4ab+ b²= (2a- b)²
в) 100p² -60p+ 9= (10p- 3)²
г) 49x²y² - 14xy+ 1= (7xy- 1)²
д) t⁴  -26t² + 169= (t² - 13)²

Answer 3

а) m^2 + 2mn + n^2

To express this polynomial as the square of a binomial, we need to find two terms that, when squared, give us the original polynomial.

In this case, we can see that the first term, m^2, is the square of m, and the last term, n^2, is the square of n. So, we can write the given polynomial as:

(m + n)^2

б) 4a^2 - 4ab + b^2

To express this polynomial as the square of a binomial, we need to find two terms that, when squared, give us the original polynomial.

In this case, the first term, 4a^2, is the square of 2a, and the last term, b^2, is the square of b. So, we can write the given polynomial as:

(2a - b)^2

в) 100 + p^2 - 60p + 9

To express this polynomial as the square of a binomial, we need to find two terms that, when squared, give us the original polynomial.

In this case, the first term, 100, is the square of 10, and the last term, 9, is the square of 3. However, we need to consider the middle term, -60p, which is not a perfect square.

To proceed, we can try to factorize the middle term. The coefficient -60 can be factored as -6 * 10, and the variable p can be factored as p * 1. So, we can rewrite the middle term as -6p * 10.

Now, we can rewrite the polynomial as:

(10 + p)^2 - 6p * 10

Simplifying further, we get:

(10 + p)^2 - 60p

So, the given polynomial can be expressed as the square of the binomial (10 + p) minus 60p.

г) 49x^2 + y^2 - 14xy + 1

To express this polynomial as the square of a binomial, we need to find two terms that, when squared, give us the original polynomial.

In this case, the first term, 49x^2, is the square of 7x, and the last term, 1, is the square of 1. However, we need to consider the middle term, -14xy, which is not a perfect square.

To proceed, we can try to factorize the middle term. The coefficient -14 can be factored as -7 * 2, and the variables x and y can be factored as x * y. So, we can rewrite the middle term as -7xy * 2.

Now, we can rewrite the polynomial as:

(7x - y)^2 - 7xy * 2 + 1

So, the given polynomial can be expressed as the square of the binomial (7x - y) minus 7xy times 2 plus 1.

д) t^4 - 26t^2 + 169

To express this polynomial as the square of a binomial, we need to find two terms that, when squared, give us the original polynomial.

In this case, the first term, t^4, is the square of t^2, and the last term, 169, is the square of 13. However, we need to consider the middle term, -26t^2, which is not a perfect square.

To proceed, we can try to factorize the middle term. The coefficient -26 can be factored as -13 * 2, and the variable t^2 can be factored as t * t. So, we can rewrite the middle term as -13t^2 * 2.

Now, we can rewrite the polynomial as:

(t^2 - 13)^2 - 13t^2 * 2

Simplifying further, we get:

(t^2 - 13)^2 - 26t^2

So, the given polynomial can be expressed as the square of the binomial (t^2 - 13) minus 26t^2.

I hope this helps! Let me know if you have any further questions.