Answer :
Certainly! Let's work through the factorization step-by-step.
Given expression:
[tex]\[ 2x^2 - xy - 6y^2 \][/tex]
We need to factorize this quadratic expression. To do this, we look for two binomials whose product equals the given expression. Let's denote these binomials as:
[tex]\[ (Ax + By)(Cx + Dy) \][/tex]
First, we expand the product of these two binomials:
[tex]\[ (Ax + By)(Cx + Dy) = ACx^2 + ADxy + BCxy + BDy^2 \][/tex]
Combining like terms:
[tex]\[ (Ax + By)(Cx + Dy) = ACx^2 + (AD + BC)xy + BDy^2 \][/tex]
Now, compare this expanded form with our given expression [tex]\( 2x^2 - xy - 6y^2 \)[/tex]:
Matching coefficients, we get:
[tex]\[ \begin{cases} AC = 2 \\ AD + BC = -1 \\ BD = -6 \end{cases} \][/tex]
We need to find values of [tex]\( A, B, C, \)[/tex] and [tex]\( D \)[/tex] that satisfy these conditions.
Let's try to find suitable values step-by-step:
1. Find [tex]\( A \)[/tex] and [tex]\( C\)[/tex]:
[tex]\[ AC = 2 \][/tex]
Possible pairs [tex]\((A, C)\)[/tex] could be [tex]\((1, 2)\)[/tex], [tex]\((2, 1)\)[/tex], [tex]\((-1, -2)\)[/tex], [tex]\((-2, -1)\)[/tex].
2. Find [tex]\( B \)[/tex] and [tex]\( D\)[/tex]:
[tex]\[ BD = -6 \][/tex]
Possible pairs [tex]\((B, D)\)[/tex] could be [tex]\((1, -6)\)[/tex], [tex]\((-1, 6)\)[/tex], [tex]\((2, -3)\)[/tex], [tex]\((-2, 3)\)[/tex], [tex]\((3, -2)\)[/tex], [tex]\((-3, 2)\)[/tex], [tex]\((6, -1)\)[/tex], [tex]\((-6, 1)\)[/tex].
We substitute these pairs into the second equation ([tex]\(AD + BC = -1\)[/tex]) to check which pair works.
Check pair [tex]\( (A = 1, C = 2) \)[/tex]:
- If [tex]\(A = 1\)[/tex] and [tex]\(C = 2\)[/tex], then [tex]\( AD + BC = 1\cdot D + 2\cdot B = -1 \)[/tex]:
- Try [tex]\( B = 3, D = -2\)[/tex]:
[tex]\[ 1\cdot (-2) + 2\cdot 3 = -2 + 6 = 4 \quad \text{(not correct)}\][/tex]
- Try [tex]\( B = -3, D = 2 \)[/tex]:
[tex]\[ 1\cdot 2 + 2\cdot (-3) = 2 - 6 = -4 \quad \text{(not correct)}\][/tex]
Check pair [tex]\( (A = 2, C = 1) \)[/tex]:
- If [tex]\(A = 2\)[/tex] and [tex]\(C = 1\)[/tex], then [tex]\( AD + BC = 2\cdot D + 1\cdot B = -1 \)[/tex]:
- Try [tex]\( B = -3, D = 1 \)[/tex]:
[tex]\[ 2\cdot 1 + 1\cdot (-3) = 2 - 3 = -1 \quad \text{(correct)}\][/tex]
Since pair [tex]\( (A = 2, D = 1)\)[/tex] and [tex]\( (C = 1, B = -3)\)[/tex] satisfy the conditions, our factors are:
[tex]\[ (2x + 3y)(x - 2y) \][/tex]
Thus, the factorized form of the given expression is:
[tex]\[ 2x^2 - xy - 6y^2 = (x - 2y)(2x + 3y) \][/tex]
Given expression:
[tex]\[ 2x^2 - xy - 6y^2 \][/tex]
We need to factorize this quadratic expression. To do this, we look for two binomials whose product equals the given expression. Let's denote these binomials as:
[tex]\[ (Ax + By)(Cx + Dy) \][/tex]
First, we expand the product of these two binomials:
[tex]\[ (Ax + By)(Cx + Dy) = ACx^2 + ADxy + BCxy + BDy^2 \][/tex]
Combining like terms:
[tex]\[ (Ax + By)(Cx + Dy) = ACx^2 + (AD + BC)xy + BDy^2 \][/tex]
Now, compare this expanded form with our given expression [tex]\( 2x^2 - xy - 6y^2 \)[/tex]:
Matching coefficients, we get:
[tex]\[ \begin{cases} AC = 2 \\ AD + BC = -1 \\ BD = -6 \end{cases} \][/tex]
We need to find values of [tex]\( A, B, C, \)[/tex] and [tex]\( D \)[/tex] that satisfy these conditions.
Let's try to find suitable values step-by-step:
1. Find [tex]\( A \)[/tex] and [tex]\( C\)[/tex]:
[tex]\[ AC = 2 \][/tex]
Possible pairs [tex]\((A, C)\)[/tex] could be [tex]\((1, 2)\)[/tex], [tex]\((2, 1)\)[/tex], [tex]\((-1, -2)\)[/tex], [tex]\((-2, -1)\)[/tex].
2. Find [tex]\( B \)[/tex] and [tex]\( D\)[/tex]:
[tex]\[ BD = -6 \][/tex]
Possible pairs [tex]\((B, D)\)[/tex] could be [tex]\((1, -6)\)[/tex], [tex]\((-1, 6)\)[/tex], [tex]\((2, -3)\)[/tex], [tex]\((-2, 3)\)[/tex], [tex]\((3, -2)\)[/tex], [tex]\((-3, 2)\)[/tex], [tex]\((6, -1)\)[/tex], [tex]\((-6, 1)\)[/tex].
We substitute these pairs into the second equation ([tex]\(AD + BC = -1\)[/tex]) to check which pair works.
Check pair [tex]\( (A = 1, C = 2) \)[/tex]:
- If [tex]\(A = 1\)[/tex] and [tex]\(C = 2\)[/tex], then [tex]\( AD + BC = 1\cdot D + 2\cdot B = -1 \)[/tex]:
- Try [tex]\( B = 3, D = -2\)[/tex]:
[tex]\[ 1\cdot (-2) + 2\cdot 3 = -2 + 6 = 4 \quad \text{(not correct)}\][/tex]
- Try [tex]\( B = -3, D = 2 \)[/tex]:
[tex]\[ 1\cdot 2 + 2\cdot (-3) = 2 - 6 = -4 \quad \text{(not correct)}\][/tex]
Check pair [tex]\( (A = 2, C = 1) \)[/tex]:
- If [tex]\(A = 2\)[/tex] and [tex]\(C = 1\)[/tex], then [tex]\( AD + BC = 2\cdot D + 1\cdot B = -1 \)[/tex]:
- Try [tex]\( B = -3, D = 1 \)[/tex]:
[tex]\[ 2\cdot 1 + 1\cdot (-3) = 2 - 3 = -1 \quad \text{(correct)}\][/tex]
Since pair [tex]\( (A = 2, D = 1)\)[/tex] and [tex]\( (C = 1, B = -3)\)[/tex] satisfy the conditions, our factors are:
[tex]\[ (2x + 3y)(x - 2y) \][/tex]
Thus, the factorized form of the given expression is:
[tex]\[ 2x^2 - xy - 6y^2 = (x - 2y)(2x + 3y) \][/tex]
