Juan Hewlett

2021-12-28

To determine: Factor

a)${x}^{2}-4x+3$

b)$2{y}^{2}-5y+2$

c)$6{z}^{2}-13z+6$

a)

b)

c)

abonirali59

Beginner2021-12-29Added 35 answers

Step 1

a) Factoring Quadratics:

We can get a quadraric by multiplying two first-degree polynomials

For example:

$(x+4)(x-2)$

By using FOIL method

$(x+4)(x-2)=x(x-2)+4(x-2)$

$={x}^{2}-2x+4x-8$

$={x}^{2}+2x-8$

Factoring quadratics needs FOIL backward because factoring is the reverse of multiplication

Given:${x}^{2}-4x+3$

We have to find integers b and d such that

${x}^{2}-4x+3=(x+b)(x+d)$

$={x}^{2}+dx+bx+bd$

${x}^{2}-4x+3={x}^{2}+(b+d)x+bd$

Since the constant coefficients on each side of the equation ought to be equal, we must have$bd=3$ (i.e) b and d are factors of 3.

Similarly, the coefficients of x must be the same, so that$b+d=-4$

The following table shows the possibilities

$$\begin{array}{|cc|}\hline \text{Factors b, d of 3}& \text{Sum}\text{}b+d=-4\\ 1\times 3& 1+3=4\\ 1(-3)& 1-3=-2\\ (-1)(-3)& -1-3=-4\\ \hline\end{array}$$

From the above table, the only factors with product 3 and the sum -4 are -1 and -3

So the correct factorization os${x}^{2}-4x+3=(x-1)(x-3)$

To check:

$(x-1)(x-3)=x(x-3)-1(x-3)$

$={x}^{2}-3x-x+3$

$={x}^{2}-4x+3$

a) Factoring Quadratics:

We can get a quadraric by multiplying two first-degree polynomials

For example:

By using FOIL method

Factoring quadratics needs FOIL backward because factoring is the reverse of multiplication

Given:

We have to find integers b and d such that

Since the constant coefficients on each side of the equation ought to be equal, we must have

Similarly, the coefficients of x must be the same, so that

The following table shows the possibilities

From the above table, the only factors with product 3 and the sum -4 are -1 and -3

So the correct factorization os

To check:

Juan Spiller

Beginner2021-12-30Added 38 answers

Step 1

b) Given:$2{y}^{2}-5y+2$

We have to find integers a, b, c and d such that

$2{y}^{2}-5y+2=(ay+b)(cy+d)$

$=ac{y}^{2}+ady+bcy+bd$

$2{y}^{2}-55y+2=ac{y}^{2}+(ad+bc)y+bd$

Since the coefficient of$y}^{2$ ought to be same on both sides, we find that $ac=2$ . Likewise, the constant term $bd=2$ . The positive factors of 2 are 2 and 1. Since the midterm is negative, we consider only negative factors of 2. The possibilities are -2 and -1. Now we have to try various arrangements of these factors until we find the one which gives correct coefficient of y.

$(2y-2)(y-1)=2y(y-1)-2(y-1)$

$=2{y}^{2}-2y-2y+2$

$=2{y}^{2}-4y+2$

$(2y-1)(y-2)=2y(y-2)-1(y-2)$

$=2{y}^{2}-4y-y+2$

$=2{y}^{2}-5y+2$

The last trial gives the correct factorization.

b) Given:

We have to find integers a, b, c and d such that

Since the coefficient of

The last trial gives the correct factorization.

karton

Expert2022-01-05Added 613 answers

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