Question

Factorize by using middle term splitting 1). Y^2-11y+28​

Answer 1

Answer:

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Step-by-step explanation:

y2-11y+28=0

Two solutions were found :

y = 7

y = 4

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "y2" was replaced by "y^2".

Step by step solution :

Step 1 :

Trying to factor by splitting the middle term

1.1 Factoring y2-11y+28

The first term is, y2 its coefficient is 1 .

The middle term is, -11y its coefficient is -11 .

The last term, "the constant", is +28

Step-1 : Multiply the coefficient of the first term by the constant 1 • 28 = 28

Step-2 : Find two factors of 28 whose sum equals the coefficient of the middle term, which is -11 .

-28 + -1 = -29

-14 + -2 = -16

-7 + -4 = -11 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -7 and -4

y2 - 7y - 4y - 28

Step-4 : Add up the first 2 terms, pulling out like factors :

y • (y-7)

Add up the last 2 terms, pulling out common factors :

4 • (y-7)

Step-5 : Add up the four terms of step 4 :

(y-4) • (y-7)

Which is the desired factorization

Equation at the end of step 1 :

(y - 4) • (y - 7) = 0

Step 2 :

Theory - Roots of a product :

2.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

2.2 Solve : y-4 = 0

Answer 2

The factors are (y - 7)(y - 4).

Step-by-step explanation:

The expression is:

$y^{2}-11y+28$

Factorize the expression by splitting the middle term as follows:

$y^{2}-11y+28$

$=y^{2}-7y-4y+28\\=y(y-7)-4(y-7)\\=(y-7)(y-4)$

Thus, the factors are (y - 7)(y - 4).

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