How to factor a polynomial by grouping? Questions with detailed solutions and explanations are presented.
Factor completely the polynomial \[ 4x^2 + 4x + 3x + 3 \]
Note that all four terms in the given polynomial have no common factor. However, by grouping the first two terms, we can factor \( \color{red} {4x} \) out as follows:
\[ 4x^2 + 4x = 4x(x + 1) \]We now group the last two terms and factor \( \color{red} 3 \) out as follows:
\[ 3x + 3 = 3(x + 1) \]Rewrite the given polynomial with the grouped terms in factored form:
\[ 4x^2 + 4x + 3x + 3 = 4x\, {(x + 1)} + 3\, {(x + 1)} \]Note that \( \color{red} {(x + 1)} \) is a common factor which can be factored out as follows:
\[ 4x^2 + 4x + 3x + 3 = 4x\, {(x + 1)} + 3\, {(x + 1)} = (x + 1)(4x + 3) \]Factor the polynomial \[ 2x^2 - 4x + 3xy - 6y \]
There is no common factor to all four terms in the given polynomial.
Group the first two terms and factor \( 2x \) out:
\[ 2x^2 - 4x = 2x(x - 2) \]Group the last two terms and factor \( 3y \) out:
\[ 3xy - 6y = 3y(x - 2) \]Rewrite the given polynomial as follows:
\[ 2x^2 - 4x + 3xy - 6y = 2x\, {(x - 2)} + 3y\,{(x - 2)} \]Factor \( x - 2 \) out and rewrite the given polynomial in factored form as follows:
\[ 2x^2 - 4x + 3xy - 6y = (x -2) (2x + 3y) \]
The terms in the given polynomial have no common factor.
The first two terms can be grouped and factored as follows, factoring \( x \) out:
\[ xy - x = x(y - 1) \]The last two terms can be factored by taking \( 2 \) out:
\[ -2y + 2 = 2(-y + 1) = -2(y - 1) \]Rewrite the given polynomial in factored form as follows:
\[ xy - x - 2y + 2 = x\,\color{red}{(y - 1)} - 2\,\color{red}{(y - 1)} \]Factor out the common factor \( \color{red}{(y - 1)} \) to factor completely:
\[ xy - x - 2y + 2 = (y - 1)(x - 2) \]
Factor completely the polynomial \[ 3x^2 + 4x + 1 \]
There is no common factor to the terms in the given polynomial. One way is to rewrite the polynomial with four terms that may be factored by grouping.
We use the identity \( 4x = 3x + x \) to rewrite the given polynomial as follows:
\[ 3x^2 + 4x + 1 = 3x^2 + 3x + x + 1 \]We group the first two terms and factor \( 3x \) out as follows:
\[ 3x^2 + 3x = 3x(x + 1) \]Rewrite the given polynomial with the grouped terms in factored form:
\[ 3x^2 + 4x + 1 = 3x(x + 1) + 1(x + 1) \]Note that \( \color{red}{(x + 1)} \) is a common factor which can be factored out as follows:
\[ 3x^2 + 4x + 1 = (x + 1)(3x + 1) \]Use grouping to factor the following polynomials completely
We first find a common factor in \( 2x^2 - 4x \) and factor it as follows:
\[ 2x^2 - 4x = 2x(x - 2) \]Next, we find a common factor in \( xy - 2y \) and factor it as follows:
\[ xy - 2y = y(x - 2) \]Now use the common factor \( (x - 2) \) and factor the given polynomial as follows:
\[ 2x^2 - 4x + xy - 2y = (2x^2 - 4x) + (xy - 2y) \] \[ = 2x(x - 2) + y(x - 2) \] \[ = (x - 2)(2x + y) \]Find a common factor in \( x^2 + 3x \) and factor it as follows:
\[ x^2 + 3x = x(x + 3) \]Next, find a common factor in \( -2x - 6 \) and factor it:
\[ -2x - 6 = -2(x + 3) \]Use the common factor \( (x + 3) \) to factor the given polynomial:
\[ x^2 + 3x - 2x - 6 = (x^2 + 3x) + (-2x - 6) \] \[ = x(x + 3) - 2(x + 3) \] \[ = (x + 3)(x - 2) \]The graph of the polynomial in part (b),
\[ y = x^2 + 3x - 2x - 6 \]is shown below. The values \( x = -3 \) and \( x = 2 \) make the factors \( (x + 3) \) and \( (x - 2) \) equal to zero, respectively. Therefore, \( x = -3 \) and \( x = 2 \) are the x-intercepts of the graph of the polynomial.
Conclusion: One way to check our factoring is to graph the given polynomial and check that the x intercepts corresponds to the zeros of the factors included in the factorization.
Find a common factor in \( 15x^2 - 3x \) and factor it as follows:
\[ 15x^2 - 3x = 3x(5x - 1) \]Next, find a common factor in \( 10x - 2 \) and factor it:
\[ 10x - 2 = 2(5x - 1) \]Use the common factor \( (5x - 1) \) to factor the given polynomial completely:
\[ 15x^2 - 3x + 10x - 2 = (15x^2 - 3x) + (10x - 2) \] \[ = 3x(5x - 1) + 2(5x - 1) \] \[ = (5x - 1)(3x + 2) \]The given polynomial has three terms with no common factor. One way to factor is to rewrite it by replacing \( x \) with \( 4x - 3x \) as follows:
\[ 4x^2 + x - 3 = 4x^2 + 4x - 3x - 3 \]We can now factor \( 4x^2 + 4x \) as follows:
\[ 4x^2 + 4x = 4x(x + 1) \]Next, factor \( -3x - 3 \) as follows:
\[ -3x - 3 = -3(x + 1) \]Use the common factor \( (x + 1) \) to factor the given polynomial completely:
\[ 4x^2 + x - 3 = 4x^2 + 4x - 3x - 3 = (4x^2 + 4x) + (-3x - 3) \] \[ = 4x(x + 1) - 3(x + 1) = (x + 1)(4x - 3) \]Note that \( x \) is a common factor to all terms in the given polynomial. Hence, we start by factoring as follows:
\[ x^2y + 3x + x^2y^2 + 3xy = x(xy + 3 + xy^2 + 3y) \]Rewrite by grouping terms as follows:
\[ x^2y + 3x + x^2y^2 + 3xy = x((xy + xy^2) + (3 + 3y)) \]The terms in \( (xy + xy^2) \) have the factor \( xy \), and the terms in \( (3 + 3y) \) have the common factor \( 3 \). Hence, we factor as follows:
\[ x^2y + 3x + x^2y^2 + 3xy = x((xy + xy^2) + (3 + 3y)) \] \[ = x(xy(1 + y) + 3(1 + y)) = x(1 + y)(xy + 3) \]Note that there are 5 terms in the given polynomial with a common factor to all of them. Rewrite the polynomial by replacing \( -x \) with \( -3x + 2x \) as follows:
\[ 3x^2 + 3xy - x + 2y - 2 = 3x^2 + 3xy - 3x + 2x + 2y - 2 \]We shall now factor the equivalent polynomial \( 3x^2 + 3xy - 3x + 2x + 2y - 2 \). We can group the first 3 terms and factor as follows:
\[ 3x^2 + 3xy - 3x = 3x(x + y - 1) \]We now group the last three terms and factor as follows:
\[ 2x + 2y - 2 = 2(x + y - 1) \]The two groups have the common factor \( (x + y - 1) \), and the given polynomial is factored as follows:
\[ = (3x^2 + 3xy - 3x) + (2x + 2y - 2) \] \[ = 3x(x + y - 1) + 2(x + y - 1) = (x + y - 1)(3x + 2) \]