Factoring Trinomials

-3

You need two numbers that multiply to −6 and add to −1: 2 and −3. So $d^2 - d - 6 = (d + 2)(d - 3)$, and $k = -3$.

3: Check the middle term: $(d + 2)(d + 3) = d^2 + 5d + 6$. You need a product of −6 and a sum of −1.

$5x - 2$

$a \cdot c = 60$; find factors of 60 that add to −19: −15 and −4. Rewrite: $10x^2 - 15x - 4x + 6 = 5x(2x - 3) - 2(2x - 3) = (5x - 2)(2x - 3)$.

$10x - 3$: The factors of 6 were chosen without checking that the middle term comes out to −19x.

$10x - 1$: The factors of 6 were chosen without checking that the middle term comes out to −19x.

$5x - 3$: The grouping was right, but the constants were switched when factoring out the common factor.

$k(x) = (x + 5)(x - 3)$

You need two numbers that multiply to −15 and add to +2: 5 and −3. So $x^2 + 2x - 15 = (x + 5)(x - 3)$.

$k(x) = (x + 15)(x - 1)$: 15 and −1 multiply to −15 but add to 14, not 2. Check the middle term.

$k(x) = (x + 1)(x - 15)$: 1 and −15 multiply to −15 but add to −14, not 2. Check the middle term.

$k(x) = (x + 3)(x - 5)$: 3 and −5 multiply to −15 but add to −2. The signs are switched.

$g(x) = (x + 18)(x - 3)$

You need two numbers that multiply to −54 and add to 15: 18 and −3. Check: $18 \cdot (-3) = -54$ and $18 + (-3) = 15$.

$g(x) = (x + 9)(x - 6)$: 9 and −6 multiply to −54, but the middle term would be $3x$, not $15x$.

$g(x) = (x - 9)(x + 6)$: −9 and 6 multiply to −54, but the middle term would be $-3x$, not $15x$.

$g(x) = (x - 18)(x + 3)$: The middle term would be $-15x$. The signs are switched.

$(12x - 5)(2x - 1)$

Multiply out: $12x \cdot 2x = 24x^2$, the middle terms are $-12x - 10x = -22x$, and $(-5)(-1) = 5$.

$(12x + 5)(2x + 1)$: The first and last terms match, but the middle term is $+22x$, not $-22x$.

$(8x - 5)(3x - 1)$: The first and last terms match, but the middle term is $-23x$.

$(8x + 5)(3x + 1)$: The first and last terms match, but the middle term is $+23x$.

$q(x) = (3x - 4)^2$

This is a perfect square trinomial: $9x^2 = (3x)^2$, $16 = 4^2$, and $2(3x)(-4) = -24x$. So $q(x) = (3x - 4)^2$.

$q(x) = (9x - 4)(x - 4)$: The middle term would be $-36x - 4x = -40x$, not $-24x$.

$q(x) = (3x + 4)^2$: The middle term would be $+24x$. The sign must be negative.

$q(x) = (9x + 4)(x + 4)$: The middle term would be $+40x$, not $-24x$.

$(x-9)(2x-7)$

$(x-9)(2x-7)=2x^2-7x-18x+63=2x^2-25x+63$.

$(x+9)(2x+7)$: The middle terms are positive, producing $+25x$.

$(x+7)(2x+9)$: This gives positive middle terms and the wrong coefficient combination.

$(x-7)(2x-9)$: This expands to $2x^2-23x+63$, not $-25x$.

$(4w+3)^2$

This is a perfect square: $(4w)^2+2(4w)(3)+3^2=(4w+3)^2$.

$(4w-3)^2$: This gives middle term $-24w$.

$(8w+3)^2$: This gives $64w^2+48w+9$.

$(8w-3)^2$: This gives $64w^2-48w+9$.

$(x+4)(x+6)$

4 and 6 multiply to 24 and add to 10, so $x^2+10x+24=(x+4)(x+6)$.

$(x+1)(x+24)$: The constants multiply to 24 but add to 25.

$(x+2)(x+12)$: The constants add to 14, not 10.

$(x+3)(x+8)$: The constants add to 11, not 10.

$3(2x^2 + x - 3)$ and $3(2x + 3)(x - 1)$

Factor out the greatest common factor, 3, to get $3(2x^2 + x - 3)$. Factor the trinomial to get $3(2x + 3)(x - 1)$.

$-3(2x^2 + x + 3)$: The sign was changed when factoring out the greatest common factor.

$3(2x - 3)(x + 1)$: These binomial factors do not produce the required linear term, $3x$.

$-3(2x - 3)(x - 1)$: The signs are incorrect when the expression is expanded.

$12(x - 2)^2$ and $12(x^2 - 4x + 4)$

Factor out 12 to get $12(x^2 - 4x + 4)$. The trinomial is a perfect square, so it also equals $12(x - 2)^2$.

$-12(x^2 + 4x + 4)$: The negative factor changes the leading and constant terms to negative values.

$12(x - 4)(x - 1)$: These factors produce a linear term of $-60x$, not $-48x$.

$-12(x + 2)^2$: Expanding gives $-12x^2 - 48x - 48$, with incorrect leading and constant terms.

$x - 8$

Factor out 5: $5(x^2 - 6x - 16)$. The numbers −8 and 2 have a product of −16 and a sum of −6, so the expression is $5(x - 8)(x + 2)$. Therefore $x - 8$ is a factor.

$x + 4$: −4 and 4 multiply to −16, but they do not add to −6, the linear coefficient.

$x + 8$: A sign error gave $5(x + 8)(x - 2)$.

$x - 4$: −4 and 4 multiply to −16, but they do not add to −6, the linear coefficient.

$x - 4$

Factor out 2: $2(x^2 - 8x + 16)$. The numbers −4 and −4 have a sum of −8 and a product of 16, so the expression is $2(x - 4)(x - 4)$, and $x - 4$ is a factor.

$x + 4$: 4 and 4 multiply to 16, but they add to 8, not −8.

$2x - 1$: −1 and −32 multiply to 32, but $(2x - 1)(x - 32)$ does not produce the linear term $-16x$.

$x + 2$: 2 and 8 multiply to 16, but they do not add to −8.

$(3x - 7)(x + 6)$

Multiply $3 \cdot (-42) = -126$. The terms $18x$ and $-7x$ have a product of $-126x^2$ and a sum of $11x$. Then $3x^2 + 18x - 7x - 42 = 3x(x + 6) - 7(x + 6) = (3x - 7)(x + 6)$.

$(3x - 7)(x - 6)$: The expression was expanded correctly, but sign errors were made when factoring each group.

$(3x + 7)(x + 6)$: The linear term and the sign of the constant were not considered.

None of these: The expression is factorable; a leading coefficient other than 1 can be handled by grouping.