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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.

اب آپ هيک کليچر اس کو حل نہیں کر سکتا ، ليکن اس کے ٹکڑے حسابي هے. نيچے سے کو ئي کوشش کريں يا خود کو ٹائپ کريں

تربيت (40)

هر کيس پر اول آزمائيں جواب کو چيک کر نے کے ليے ظاہر کريں جو صحيح هے وہ حل کر نے کے ليے کل قدم کے ليے کھولي جا سکتے هيں

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

    جواب کھوليں

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

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

    جواب کھوليں

    ⓐ −15 ⓑ −3

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

    جواب کھوليں

    ⓐ −54 ⓑ 54

  4. Simplify: ⓐ \(|-5|\) ⓑ \(|3|.\)

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

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

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

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

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

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

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

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

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

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

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

    \({y}^{2}-6y+15\)
    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: \({m}^{2}+4m+18\).

    جواب کھوليں

    prime

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

    جواب کھوليں

    prime

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

    جواب کھوليں

    \(2x+{x}^{2}-48\)
    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: \(9m+{m}^{2}+18\).

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

    \({x}^{2}+12xy+36{y}^{2}\)
    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: \({u}^{2}+11uv+28{v}^{2}\).

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

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

    جواب کھوليں

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

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

    جواب کھوليں

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

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

    جواب کھوليں

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

    جواب کھوليں

    prime

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

    جواب کھوليں

    prime

آپ کے اپنے کام رکھیں

مفت اکاؤنٹ تمام درسوں پر نوٹ شامل کر تا هے ، آپ نے جو ختم کر يا هے اس کا ريکورڈ ، آپ کے حل کيے گئے مسئلے ایک جگہ پر اور آپ اس صفحے کے بارے ميں سوال کر سکتے هيں تو تيرچر ريتھ ميں خود تمام لوگوں کے ليے کھلا هے ، خواہ آپ داخل هے يا نہيں

رجسٹر لاگ ان

علامتیں

تمام تعريف کے ليے ، تصوير کے ليے اور اس ميں ہر حروف کا مطلب کے ليے کو ئي علامت کو ٹک کريں

کیسے: 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

سوالات لوگ پوچھتے ہیں

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.

اس صفحے کے حصے سے تناسب میں لایا گیا ہے OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0). یہاں سے کڑوا اور دوبارہ بیان کیا گیا ہے؛ غلطیاں ہماری ہیں. اسکے ذريعے کو غير تجاري استعمال کے ليے لائسنس ديے گيا هے اس ليکھے کو مفت ميں ملتا هے اور اسکے ليے کوئی خرچ نہیں کريے جا تا هے اور اس کو عين لائسنس کے تحت شيڑيا جا تا هے

میں زیادہ Algebra

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