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Factor Trinomials of the Form x

Factor trinomials of the form

Factor Trinomials of the Form

You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication, to start with the product and end up with the factors. Let’s look at an example of multiplying binomials to refresh your memory.

To factor the trinomial means to start with the product, \({x}^{2}+5x+6\), and end with the factors, \((x+2)(x+3)\). You need to think about where each of the terms in the trinomial came from.

The first term came from multiplying the first term in each binomial. So to get \({x}^{2}\) in the product, each binomial must start with an x.

\[\begin{array}{l}{x}^{2}+5x+6 \\ (x\ )(x\ )\end{array}\]

The last term in the trinomial came from multiplying the last term in each binomial. So the last terms must multiply to 6.

What two numbers multiply to 6?

The factors of 6 could be 1 and 6, or 2 and 3. How do you know which pair to use?

Consider the middle term. It came from adding the outer and inner terms.

Factors of \(6\)Sum of factors
\(1,6\)\(1+6=7\)
\(2,3\)\(2+3=5\)
\[\begin{array}{llll}{x}^{2}+5x+6 & & & \text{product} \\ (x+2)(x+3) & & & \text{factors}\end{array}\]\[\begin{array}{llll}{x}^{2}+5x+6 & & & {x}^{2}+bx+c \\ (x+2)(x+3) & & & (x+m)(x+n)\end{array}\]
How to Factor Trinomials of the Form

Try it.

Factor: \({x}^{2}+7x+12\).

Solution
Example

Try it.

Factor: \({u}^{2}+11u+24\).

Solution

Notice that the variable is u, so the factors will have first terms u.

\(\begin{array}{llll} & & & {u}^{2}+11u+24 \\ \text{Write the factors as two binomials with first terms}\ u. & & & (u\ )(u\ )\end{array}\)

Find two numbers that: multiply to 24 and add to 11.

Factors of \(24\)Sum of factors
\(\ 1,24\)\(1+24=25\)
\(\ 2,12\)\(2+12=14\)
\(3,8\)\(\ 3+8=11\text{*}\)
\(4,6\)\(\ 4+6=10\)

\(\begin{array}{llll}\text{Use 3 and 8 as the last terms of the binomials.} & & & \ (u+3)(u+8) \\ \\ \text{Check.} & & & \\ \\ \\ (u+3)(u+8) & & & \\ {u}^{2}+3u+8u+24 & & & \\ {u}^{2}+11u+24\ ✓ & & & \end{array}\)

Condensed: the full section is in OpenStax Elementary Algebra 2e.

Factor Trinomials of the Form

Sometimes you’ll need to factor trinomials of the form \({x}^{2}+bxy+c{y}^{2}\) with two variables, such as \({x}^{2}+12xy+36{y}^{2}.\) The first term, \({x}^{2}\), is the product of the first terms of the binomial factors, \(x\cdot x\). The \({y}^{2}\) in the last term means that the second terms of the binomial factors must each contain y. To get the coefficients b and c, you use the same process summarized in the previous objective.

Example

Try it.

Factor: \({x}^{2}+12xy+36{y}^{2}\).

Solution

\({x}^{2}+12xy+36{y}^{2}\)
Note that the first terms are x, last terms contain y.\((x_y)(x_y)\)

Find the numbers that multiply to 36 and add to 12.

Factors of \(36\)Sum of factors
1, 36\(1+36=37\)
2, 18\(2+18=20\)
3, 12\(3+12=15\)
4, 9\(4+9=13\)
6, 6\(6+6=12\text{*}\)

Use 6 and 6 as the coefficients of the last terms.\((x+6y)(x+6y)\)
Check your answer.
\(\begin{array}{l} \\ \\ (x+6y)(x+6y) \\ {x}^{2}+6xy+6xy+36{y}^{2} \\ {x}^{2}+12xy+36{y}^{2}✓\end{array}\)

Example

Try it.

Factor: \({r}^{2}-8rs-9{s}^{2}\).

Solution

We need \(r\) in the first term of each binomial and \(s\) in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

\(\begin{array}{llll} & & & {r}^{2}-8rs-9{s}^{2} \\ \text{Note that the first terms are}\ r,\ \text{last terms contain}\ s. & & & (r_s)(r_s)\end{array}\)

Find the numbers that multiply to \(-9\) and add to \(-8\).

Factors of \(-9\)Sum of factors
\(\ 1,-9\)\(\ 1+(-9)=-8\text{*}\)
\(-1,9\)\(\ -1+9=8\)
\(\ 3,-3\)\(3+(-3)=0\)

\(\begin{array}{llll}\text{Use}\ 1,-9\ \text{as coefficients of the last terms.} & & & \ (r+s)(r-9s) \\ \text{Check your answer.} & & & \\ \\ \\ (r-9s)(r+s) & & & \\ {r}^{2}+rs-9rs-9{s}^{2} & & & \\ {r}^{2}-8rs-9{s}^{2}\ ✓ & & & \end{array}\)

Example

Try it.

Factor: \({u}^{2}-9uv-12{v}^{2}\).

Solution

We need u in the first term of each binomial and \(v\) in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

\({u}^{2}-9uv-12{v}^{2}\)
Note that the first terms are u, last terms contain v.\((u_v)(u_v)\)

Find the numbers that multiply to \(-12\) and add to \(-9\).

Factors of \(-12\)Sum of factors
\(\ 1,-12\)\(1+(-12)=-11\)
\(\ -1,12\)\(-1+12=11\)
\(\ 2,-6\)\(2+(-6)=-4\)
\(-2,6\)\(-2+6=4\)
\(\ 3,-4\)\(3+(-4)=-1\)
\(-3,4\)\(-3+4=1\)

Note there are no factor pairs that give us \(-9\) as a sum. The trinomial is prime.

Key Concepts

  • Factor trinomials of the form \({x}^{2}+bx+c\)
    1. Write the factors as two binomials with first terms x: \((x\ )(x\ )\).
    2. Find two numbers m and n that
      Multiply to c, \(m\cdot n=c\)
      Add to b, \(m+n=b\)
    3. Use m and n as the last terms of the factors: \((x+m)(x+n)\).
    4. Check by multiplying the factors.

Factor Trinomials of the Form x

Factor Trinomials of the Form \({x}^{2}+bx+c\)

In the following exercises, factor each trinomial of the form \({x}^{2}+bx+c\).

Try it.

\({x}^{2}+4x+3\)

Solution

\((x+1)(x+3)\)

Try it.

\({y}^{2}+8y+7\)

Try it.

\({m}^{2}+12m+11\)

Solution

\((m+1)(m+11)\)

Try it.

\({b}^{2}+14b+13\)

Try it.

\({a}^{2}+9a+20\)

Solution

\((a+4)(a+5)\)

Try it.

\({m}^{2}+7m+12\)

Try it.

\({p}^{2}+11p+30\)

Solution

\((p+5)(p+6)\)

Try it.

\({w}^{2}+10w+21\)

Try it.

\({n}^{2}+19n+48\)

Solution

\((n+3)(n+16)\)

Try it.

\({b}^{2}+14b+48\)

Try it.

\({a}^{2}+25a+100\)

Solution

\((a+5)(a+20)\)

Try it.

\({u}^{2}+101u+100\)

Try it.

\({x}^{2}-8x+12\)

Solution

\((x-2)(x-6)\)

Try it.

\({q}^{2}-13q+36\)

Try it.

\({y}^{2}-18y+45\)

Solution

\((y-3)(y-15)\)

Try it.

\({m}^{2}-13m+30\)

Try it.

\({x}^{2}-8x+7\)

Solution

\((x-1)(x-7)\)

Try it.

\({y}^{2}-5y+6\)

Try it.

\({p}^{2}+5p-6\)

Solution

\((p-1)(p+6)\)

Try it.

\({n}^{2}+6n-7\)

Try it.

\({y}^{2}-6y-7\)

Solution

\((y+1)(y-7)\)

Try it.

\({v}^{2}-2v-3\)

Try it.

\({x}^{2}-x-12\)

Solution

\((x-4)(x+3)\)

Try it.

\({r}^{2}-2r-8\)

Try it.

\({a}^{2}-3a-28\)

Solution

\((a-7)(a+4)\)

Try it.

\({b}^{2}-13b-30\)

Try it.

\({w}^{2}-5w-36\)

Solution

\((w-9)(w+4)\)

Try it.

\({t}^{2}-3t-54\)

Try it.

\({x}^{2}+x+5\)

Solution

prime

Try it.

\({x}^{2}-3x-9\)

Try it.

\(8-6x+{x}^{2}\)

Solution

\((x-4)(x-2)\)

Try it.

\(7x+{x}^{2}+6\)

Try it.

\({x}^{2}-12-11x\)

Solution

\((x-12)(x+1)\)

Try it.

\(-11-10x+{x}^{2}\)

Factor Trinomials of the Form \({x}^{2}+bxy+c{y}^{2}\)

In the following exercises, factor each trinomial of the form \({x}^{2}+bxy+c{y}^{2}\). If the trinomial cannot be factored, answer “Prime.”

Try it.

\({p}^{2}+3pq+2{q}^{2}\)

Solution

\((p+q)(p+2q)\)

Try it.

\({m}^{2}+6mn+5{n}^{2}\)

Try it.

\({r}^{2}+15rs+36{s}^{2}\)

Solution

\((r+3s)(r+12s)\)

Try it.

\({u}^{2}+10uv+24{v}^{2}\)

Try it.

\({m}^{2}-12mn+20{n}^{2}\)

Solution

\((m-2n)(m-10n)\)

Try it.

\({p}^{2}-16pq+63{q}^{2}\)

Try it.

\({x}^{2}-2xy-80{y}^{2}\)

Solution

\((x+8y)(x-10y)\)

Try it.

\({p}^{2}-8pq-65{q}^{2}\)

Try it.

\({m}^{2}-64mn-65{n}^{2}\)

Solution

\((m+n)(m-65n)\)

Try it.

\({p}^{2}-2pq-35{q}^{2}\)

Try it.

\({a}^{2}+5ab-24{b}^{2}\)

Solution

\((a+8b)(a-3b)\)

Try it.

\({r}^{2}+3rs-28{s}^{2}\)

Try it.

\({x}^{2}-3xy-14{y}^{2}\)

Solution

prime

Try it.

\({u}^{2}-8uv-24{v}^{2}\)

Try it.

\({m}^{2}-5mn+30{n}^{2}\)

Solution

prime

Try it.

\({c}^{2}-7cd+18{d}^{2}\)

Mixed Practice

In the following exercises, factor each expression.

Try it.

\({u}^{2}-12u+36\)

Solution

\((u-6)(u-6)\)

Try it.

\({w}^{2}+4w-32\)

Try it.

\({x}^{2}-14x-32\)

Solution

\((x+2)(x-16)\)

Try it.

\({y}^{2}+41y+40\)

Try it.

\({r}^{2}-20rs+64{s}^{2}\)

Solution

\((r-4s)(r-16s)\)

Try it.

\({x}^{2}-16xy+64{y}^{2}\)

Try it.

\({k}^{2}+34k+120\)

Solution

\((k+4)(k+30)\)

Try it.

\({m}^{2}+29m+120\)

Try it.

\({y}^{2}+10y+15\)

Solution

prime

Try it.

\({z}^{2}-3z+28\)

Try it.

\({m}^{2}+mn-56{n}^{2}\)

Solution

\((m+8n)(m-7n)\)

Try it.

\({q}^{2}-29qr-96{r}^{2}\)

Try it.

\({u}^{2}-17uv+30{v}^{2}\)

Solution

\((u-15v)(u-2v)\)

Try it.

\({m}^{2}-31mn+30{n}^{2}\)

Try it.

\({c}^{2}-8cd+26{d}^{2}\)

Solution

prime

Try it.

\({r}^{2}+11rs+36{s}^{2}\)

Condensed: the full section is in OpenStax Elementary Algebra 2e.

Haddaad Ma jiro xisaabis oo ku nool midkan, laakiin qaybaha ka mid ah waa la xisaabi karaa. Ku raaxayso mid ka hooseeya, ama ku qor adiga kuu gaar ah.

Tababar (40)

Ku day mid kasta oo ka mid ah oo ku saabsan buugaag hore. Muuji jawaabta si aad u hubiso; kuwa la xaqiijiyay waxaa la furi karaa in xalka ah tallaabo kasta.

  1. Multiply: \((x+4)(x+5).\)

    Jawaabta muuji

    \({x}^{2}+9x+20\)

  2. Simplify: ⓐ \(-9+(-6)\) ⓑ \(-9+6.\)

    Jawaabta muuji

    ⓐ −15 ⓑ −3

  3. Simplify: ⓐ \(-9(6)\) ⓑ \(-9(-6).\)

    Jawaabta muuji

    ⓐ −54 ⓑ 54

  4. Simplify: ⓐ \(|-5|\) ⓑ \(|3|.\)

  5. Factor: \({x}^{2}+7x+12\).

  6. Factor: \({x}^{2}+6x+8\).

    Jawaabta muuji

    \((x+2)(x+4)\)

  7. Factor: \({y}^{2}+8y+15\).

    Jawaabta muuji

    \((y+3)(y+5)\)

  8. Factor: \({u}^{2}+11u+24\).

    Jawaabta muuji

    Notice that the variable is u, so the factors will have first terms u.

    \(\begin{array}{llll} & & & {u}^{2}+11u+24 \\ \text{Write the factors as two binomials with first terms}\ u. & & & (u\ )(u\ )\end{array}\)

    Find two numbers that: multiply to 24 and add to 11.

    Factors of \(24\)Sum of factors
    \(\ 1,24\)\(1+24=25\)
    \(\ 2,12\)\(2+12=14\)
    \(3,8\)\(\ 3+8=11\text{*}\)
    \(4,6\)\(\ 4+6=10\)

    \(\begin{array}{llll}\text{Use 3 and 8 as the last terms of the binomials.} & & & \ (u+3)(u+8) \\ \\ \text{Check.} & & & \\ \\ \\ (u+3)(u+8) & & & \\ {u}^{2}+3u+8u+24 & & & \\ {u}^{2}+11u+24\ ✓ & & & \end{array}\)

  9. Factor: \({q}^{2}+10q+24\).

    Jawaabta muuji

    \((q+4)(q+6)\)

  10. Factor: \({t}^{2}+14t+24\).

    Jawaabta muuji

    \((t+2)(t+12)\)

  11. Factor: \({y}^{2}+17y+60\).

    Jawaabta muuji

    \(\begin{array}{llll} & & & {y}^{2}+17y+60 \\ \text{Write the factors as two binomials with first terms}\ y. & & & (y\ )(y\ )\end{array}\)

    Find two numbers that multiply to 60 and add to 17.

    Factors of \(60\)Sum of factors
    \(1,60\)\(1+60=61\)
    \(2,30\)\(2+30=32\)
    \(3,20\)\(3+20=23\)
    \(4,15\)\(4+15=19\)
    \(5,12\)\(\ 5+12=17\text{*}\)
    \(6,10\)\(6+10=16\)

    \(\begin{array}{llll}\text{Use}\ 5\ \text{and}\ 12\ \text{as the last terms.} & & & \ (y+5)(y+12) \\ \\ \text{Check.} & & & \\ \\ \\ (y+5)(y+12) & & & \\ ({y}^{2}+12y+5y+60) & & & \\ ({y}^{2}+17y+60)✓ & & & \end{array}\)

  12. Factor: \({x}^{2}+19x+60\).

    Jawaabta muuji

    \((x+4)(x+15)\)

  13. Factor: \({v}^{2}+23v+60\).

    Jawaabta muuji

    \((v+3)(v+20)\)

  14. Factor: \({t}^{2}-11t+28\).

    Jawaabta muuji

    Again, with the positive last term, 28, and the negative middle term, \(-11t\), we need two negative factors. Find two numbers that multiply 28 and add to \(-11\).

    \(\begin{array}{llll} & & & {t}^{2}-11t+28 \\ \\ \text{Write the factors as two binomials with first terms}\ t. & & & (t\ )(t\ )\end{array}\)

    Find two numbers that: multiply to 28 and add to \(-11\).

    Factors of \(28\)Sum of factors
    \(\ -1,-28\)\(-1+(-28)=-29\)
    \(\ -2,-14\)\(-2+(-14)=-16\)
    \(-4,-7\)\(\ -4+(-7)=-11\text{*}\)

    \(\begin{array}{llll}\text{Use}\ -4,-7\ \text{as the last terms of the binomials.} & & & \ (t-4)(t-7) \\ \text{Check.} & & & \\ \\ \\ (t-4)(t-7) & & & \\ {t}^{2}-7t-4t+28 & & & \\ {t}^{2}-11t+28\ ✓ & & & \end{array}\)

  15. Factor: \({u}^{2}-9u+18\).

    Jawaabta muuji

    \((u-3)(u-6)\)

  16. Factor: \({y}^{2}-16y+63\).

    Jawaabta muuji

    \((y-7)(y-9)\)

  17. Factor: \({z}^{2}+4z-5\).

    Jawaabta muuji

    To get a negative last term, multiply one positive and one negative. We need factors of \(-5\) that add to positive 4.

    Factors of \(-5\)Sum of factors
    \(\ 1,-5\)\(1+(-5)=-4\)
    \(-1,5\)\(\ -1+5=4\text{*}\)

    Notice: We listed both \(1,-5\ \text{and}\ -1,5\) to make sure we got the sign of the middle term correct.

    \({z}^{2}+4z-5\)
    Factors will be two binomials with first terms z.\((z\ )(z\ )\)
    Use −1, 5 as the last terms of the binomials.\((z-1)(z+5)\)
    Check.
    \(\begin{array}{l} \\ (z-1)(z+5) \\ {z}^{2}+5z-1z-5 \\ {z}^{2}+4z-5\ ✓\end{array}\)

  18. Factor: \({h}^{2}+4h-12\).

    Jawaabta muuji

    \((h-2)(h+6)\)

  19. Factor: \({k}^{2}+k-20\).

    Jawaabta muuji

    \((k-4)(k+5)\)

  20. Factor: \({z}^{2}-4z-5\).

    Jawaabta muuji

    This time, we need factors of \(-5\) that add to \(-4\).

    Factors of \(-5\)Sum of factors
    \(\ 1,-5\)\(\ 1+(-5)=-4\text{*}\)
    \(-1,5\)\(-1+5=4\)

    \({z}^{2}-4z-5\)
    Factors will be two binomials with first terms z.\((z\ )(z\ )\)
    Use 1, −5 as the last terms of the binomials.\((z+1)(z-5)\)
    Check.
    \(\begin{array}{l} \\ (z+1)(z-5) \\ {z}^{2}-5z+1z-5 \\ {z}^{2}-4z-5\ ✓\end{array}\)

    Notice that the factors of \({z}^{2}-4z-5\) are very similar to the factors of \({z}^{2}+4z-5\). It is very important to make sure you choose the factor pair that results in the correct sign of the middle term.

  21. Factor: \({x}^{2}-4x-12\).

    Jawaabta muuji

    \((x+2)(x-6)\)

  22. Factor: \({y}^{2}-y-20\).

    Jawaabta muuji

    \((y+4)(y-5)\)

  23. Factor: \({q}^{2}-2q-15\).

    Jawaabta muuji

    \(\begin{array}{llll} & & & {q}^{2}-2q-15 \\ \text{Factors will be two binomials with first terms}\ q. & & & (q\ )(q\ ) \\ \text{You can use}\ 3,-5\ \text{as the last terms of the} & & & (q+3)(q-5) \\ \text{binomials.} & & & \end{array}\)

    Factors of \(-15\)Sum of factors
    \(\ 1,-15\)\(1+(-15)=-14\)
    \(-1,15\)\(\ -1+15=14\)
    \(\ 3,-5\)\(\ 3+(-5)=-2\text{*}\)
    \(-3,5\)\(\ -3+5=2\)

    Check.

    \(\ (q+3)(q-5)\)

    \(\ {q}^{2}-5q+3q-15\)

    \(\ {q}^{2}-2q-15\ ✓\)

  24. Factor: \({r}^{2}-3r-40\).

    Jawaabta muuji

    \((r+5)(r-8)\)

  25. Factor: \({s}^{2}-3s-10\).

    Jawaabta muuji

    \((s+2)(s-5)\)

  26. Factor: \({y}^{2}-6y+15\).

    Jawaabta muuji

    \({y}^{2}-6y+15\)
    Factors will be two binomials with first terms y.\((y\ )(y\ )\)

    Factors of 15Sum of factors
    \(-1,-15\)\(-1+(-15)=-16\)
    \(-3,-5\)\(\ -3+(-5)=-8\)

    As shown in the table, none of the factors add to \(-6\); therefore, the expression is prime.

  27. Factor: \({m}^{2}+4m+18\).

    Jawaabta muuji

    prime

  28. Factor: \({n}^{2}-10n+12\).

    Jawaabta muuji

    prime

  29. Factor: \(2x+{x}^{2}-48\).

    Jawaabta muuji

    \(2x+{x}^{2}-48\)
    First we put the terms in decreasing degree order.\({x}^{2}+2x-48\)
    Factors will be two binomials with first terms x.\((x\ )(x\ )\)

    As shown in the table, you can use \(-6,8\) as the last terms of the binomials.

    \[\ (x-6)(x+8)\]
    Factors of \(-48\)Sum of factors
    \(-1,48\)\(-1+48=47\)
    \(-2,24\)
    \(-3,16\)
    \(-4,12\)
    \(-6,8\)    		\(-2+24=22\)
    \(-3+16=13\)
    \(-4+12=8\)
    \(\ -6+8=2\)

    Check.

    \(\ (x-6)(x+8)\)

    \(\ {x}^{2}-6x+8x-48\)

    \(\ {x}^{2}+2x-48\ ✓\)

  30. Factor: \(9m+{m}^{2}+18\).

    Jawaabta muuji

    \((m+3)(m+6)\)

  31. Factor: \(-7n+12+{n}^{2}\).

    Jawaabta muuji

    \((n-3)(n-4)\)

  32. Factor: \({x}^{2}+12xy+36{y}^{2}\).

    Jawaabta muuji

    \({x}^{2}+12xy+36{y}^{2}\)
    Note that the first terms are x, last terms contain y.\((x_y)(x_y)\)

    Find the numbers that multiply to 36 and add to 12.

    Factors of \(36\)Sum of factors
    1, 36\(1+36=37\)
    2, 18\(2+18=20\)
    3, 12\(3+12=15\)
    4, 9\(4+9=13\)
    6, 6\(6+6=12\text{*}\)

    Use 6 and 6 as the coefficients of the last terms.\((x+6y)(x+6y)\)
    Check your answer.
    \(\begin{array}{l} \\ \\ (x+6y)(x+6y) \\ {x}^{2}+6xy+6xy+36{y}^{2} \\ {x}^{2}+12xy+36{y}^{2}✓\end{array}\)

  33. Factor: \({u}^{2}+11uv+28{v}^{2}\).

    Jawaabta muuji

    \((u+4v)(u+7v)\)

  34. Factor: \({x}^{2}+13xy+42{y}^{2}\).

    Jawaabta muuji

    \((x+6y)(x+7y)\)

  35. Factor: \({r}^{2}-8rs-9{s}^{2}\).

    Jawaabta muuji

    We need \(r\) in the first term of each binomial and \(s\) in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

    \(\begin{array}{llll} & & & {r}^{2}-8rs-9{s}^{2} \\ \text{Note that the first terms are}\ r,\ \text{last terms contain}\ s. & & & (r_s)(r_s)\end{array}\)

    Find the numbers that multiply to \(-9\) and add to \(-8\).

    Factors of \(-9\)Sum of factors
    \(\ 1,-9\)\(\ 1+(-9)=-8\text{*}\)
    \(-1,9\)\(\ -1+9=8\)
    \(\ 3,-3\)\(3+(-3)=0\)

    \(\begin{array}{llll}\text{Use}\ 1,-9\ \text{as coefficients of the last terms.} & & & \ (r+s)(r-9s) \\ \text{Check your answer.} & & & \\ \\ \\ (r-9s)(r+s) & & & \\ {r}^{2}+rs-9rs-9{s}^{2} & & & \\ {r}^{2}-8rs-9{s}^{2}\ ✓ & & & \end{array}\)

  36. Factor: \({a}^{2}-11ab+10{b}^{2}\).

    Jawaabta muuji

    \((a-b)(a-10b)\)

  37. Factor: \({m}^{2}-13mn+12{n}^{2}\).

    Jawaabta muuji

    \((m-n)(m-12n)\)

  38. Factor: \({u}^{2}-9uv-12{v}^{2}\).

    Jawaabta muuji

    We need u in the first term of each binomial and \(v\) in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

    \({u}^{2}-9uv-12{v}^{2}\)
    Note that the first terms are u, last terms contain v.\((u_v)(u_v)\)

    Find the numbers that multiply to \(-12\) and add to \(-9\).

    Factors of \(-12\)Sum of factors
    \(\ 1,-12\)\(1+(-12)=-11\)
    \(\ -1,12\)\(-1+12=11\)
    \(\ 2,-6\)\(2+(-6)=-4\)
    \(-2,6\)\(-2+6=4\)
    \(\ 3,-4\)\(3+(-4)=-1\)
    \(-3,4\)\(-3+4=1\)

    Note there are no factor pairs that give us \(-9\) as a sum. The trinomial is prime.

  39. Factor: \({x}^{2}-7xy-10{y}^{2}\).

    Jawaabta muuji

    prime

  40. Factor: \({p}^{2}+15pq+20{q}^{2}\).

    Jawaabta muuji

    prime

Ku hay shaqadaada

A xisaab bilaash ah oo ku daray qoraalada waxbarasho kasta, diiwaanka waxa aad dhamaatay, dhibaatooyinka aad xal u tahay meel kaliya, iyo macallinka aad weydiisan kartaa oo ku saabsan bogga. Xisaabta oo keliya waa u furan yahay qof kasta, la diiwaan geliyey ama aan.

Ku soo biir Ku soo gal

Calaamadaha la isticmaalay halkan

Taab calaamad kasta oo loogu talagalay qeexidda buuxda, sawir, iyo waxa xarfaha oo dhan ka mid ah ka dhigan tahay.

Sida loo: Factor Trinomials of the Form x

  1. Factor trinomials of the form
  2. Factor trinomials of the form
  3. Write the factors as two binomials with first terms
  4. Find two numbers
  5. Use
  6. Check by multiplying the factors.
  7. the first term,
  8. the positive last term is the product of the two last terms

Su'aalaha dadku waydiiyaan

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Qaybaha ee boggaan waxaa la beddelay ka OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0). Condensed iyo dib-oo halkan lagu sharaxay; khaladaadka waa inoo. Sababtoo ah asalka waxaa loo oggolaaday in aan loo isticmaalin ganacsi, fasalkan waa bilaash, waxba ma iibiyo, iyo waxaa lagu wadaagayaa liisan la mid ah.

In ka badan Algebra

Linear equationsQuadratic equationsSystems of equationsInequalitiesFactoringExpandingSimplifying expressionsFunctions and graphsExponential and logarithmic equationsPolynomial equationsAbsolute value