$\frac { 2 }{ 5 } { x }^{ 5 }-\frac { 6 }{ 5 } { x }^{ 4 }-\frac { 14 }{ 15 } { x }^{ 2 }=$
$2xy-4x-3y+6=$
$25{ x }^{ 2 }-1=$
$36{ x }^{ 6 }-49=$
${ 5x }^{ 2 }-10x+5=$
${ x }^{ 2 }-6x+9=$
${ x }^{ 2 }-20x+100=$
${ x }^{ 2 }+10x+25=$
${ x }^{ 2 }+14x+49=$
${ x }^{ 3 }-4{ x }^{ 2 }+4x=$
$3{ x }^{ 7 }-27x=$
${ x }^{ 2 }-11x+30=$
$3{ x }^{ 2 }-10x+3$
$2{ x }^{ 2 }-x-1$

Solution of exercise 1

${ x }^{ 3 }+{ x }^{ 2 }$

${ x }^{ 3 }+{ x }^{ 2 }={ x }^{ 2 }(x+1)$

Solution of exercise 2

$2{ x }^{ 4 }+4{ x }^{ 2 }$

$2{ x }^{ 4 }+4{ x }^{ 2 }=2{ x }^{ 2 }({ x }^{ 2 }+2)$

Solution of exercise 3

${ x }^{ 2 }-4$

${ x }^{ 2 }-{ 2 }^{ 2 }=(x-2).(x+2)$

Solution of exercise 4

${ x }^{ 4 }-16$

${ ({ x }^{ 2 }) }^{ 2 }-{ 4 }^{ 2 }=({ x }^{ 2 }-4).({ x }^{ 2 }+4)$

Solution of exercise 5

$9+6x+{ x }^{ 2 }$

${ x }^{ 2 }+6x+9$

$a{ x }^{ 2 }+bx+c$

$a=1,\quad b=6,\quad c=9$

In order to break the middle term, you need to multiply "a" with "c":

$1\times 9=9$

Now break 9 into least possible factors:

$3\times 3=9$

The next step is to check the sign of the c and in this case, it is positive this means that you need to add the factors to come up with 6:

$3+3=6$

Now your middle term is broken, you can carry forward:

${ x }^{ 2 }+x(3+3)+9$

${ x }^{ 2 }+3x+3x+9$

$x(x+3)+3(x+3)$

$(x+3)(x+3)$

${(x+3) }^{ 2 }$

Solution of exercise 6

${ x }^{ 2 }-x-6=0$

$a{ x }^{ 2 }+bx+c=0$

$a=1,\quad b=-1,\quad c=-6$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(-1)+\sqrt { { (-1 })^{ 2 }-4(1)(-6) } }{ 2(1) } ,\frac { -(-1)-\sqrt { { { (-1 }) }^{ 2 }-4(1)(-6) } }{ 2(1) }$

$x=\quad \frac { 1+\sqrt { { 1 }+24 } }{ 2 } ,\frac { 1-\sqrt { { 1 }+24 } }{ 2 }$

$x=\quad \frac { 1+\sqrt { 25 } }{ 2 } ,\frac { 1-\sqrt { 25 } }{ 2 }$

$x=\quad \frac { 1+5 }{ 2 } ,\frac { 1-5 }{ 2 }$

$x=\quad \frac { 6 }{ 2 } ,\frac { -4 }{ 2 }$

$x=3,\quad x=-2$

${ x }^{ 2 }-x-6=(x-3).(x+2)$

Solution of exercise 7

${ x }^{ 4 }-10{ x }^{ 2 }+9=0$

Assuming ${ x }^{ 2 }=t$ :

${ t }^{ 2 }-10t+9=0$

$a{ t }^{ 2 }+bt+c=0$

$a=1,\quad b=-10,\quad c=9$

$t=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$t=\quad \frac { -(-10)+\sqrt { { (-10) }^{ 2 }-4(1)(9) } }{ 2(1) } ,\frac { -(-10)-\sqrt { { (-10) }^{ 2 }-4(1)(9) } }{ 2(1) }$

$t=\quad \frac { 10+\sqrt { { 100 }-36 } }{ 2 } ,\frac { 10-\sqrt { { 100 }-36 } }{ 2 }$

$t=\quad \frac { 10+\sqrt { 64 } }{ 2 } ,\frac { 1-\sqrt { 64 } }{ 2 }$

$t=\quad \frac { 10+8 }{ 2 } ,\frac { 10-8 }{ 2 }$

$t=\quad \frac { 18 }{ 2 } ,\frac { 2 }{ 2 }$

$t=9,\quad t=1$

Putting the value of t:

${ x }^{ 2 }=9,\quad { x }^{ 2 }=1$

$x=\pm 3,\quad x=\pm 1$

${ x }^{ 4 }-10{ x }^{ 2 }+9=(x+1).(x-1).(x+3).(x-3)$

Solution of exercise 8

${ x }^{ 4 }-2{ x }^{ 2 }-3=0$

Assuming ${ x }^{ 2 }=t$ :

${ t }^{ 2 }-2t-3=0$

$a{ t }^{ 2 }+bt+c=0$

$a=1,\quad b=-2,\quad c=-3$

$t=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$t=\quad \frac { -(-2)+\sqrt { { (-2) }^{ 2 }-4(1)(-3) } }{ 2(1) } ,\frac { -(-2)-\sqrt { { (-2) }^{ 2 }-4(1)(-3) } }{ 2(1) }$

$t=\quad \frac { 2+\sqrt { 4+12 } }{ 2 } ,\frac { 2-\sqrt { 4+12 } }{ 2 }$

$t=\quad \frac { 2+\sqrt { 16 } }{ 2 } ,\frac { 2-\sqrt { 16 } }{ 2 }$

$t=\quad \frac { 2+4 }{ 2 } ,\frac { 2-4 }{ 2 }$

$t=\quad \frac { 6 }{ 2 } ,\frac { -2 }{ 2 }$

$t=3,\quad t=-1$

Putting the value of t:

${ x }^{ 2 }=3,\quad { x }^{ 2 }=-1$

$x=\pm \sqrt { 3 } ,\quad x=\pm \sqrt { -1 } \notin R$

Solution of exercise 9

$2{ x }^{ 4 }+{ x }^{ 3 }-8{ x }^{ 2 }-x+6$

$P(1)=2{ (1) }^{ 4 }+{ (1) }^{ 3 }-8{ (1) }^{ 2 }-1+6=0$

$(x-1).(2{ x }^{ 3 }+3{ x }^{ 2 }-5x-6)$

$P(1)=2{ (1) }^{ 3 }+{ 3(1) }^{ 2 }-5{ (1) }-6\neq 0$

$P(-1)=2{ (-1) }^{ 3 }+{ 3(-1) }^{ 2 }-5{ (-1) }-6=-2+3+5-6=0$

$(x-1)(x+1)(2{ x }^{ 2 }+x-6)$

$P(2)=2{ (2) }^{ 2 }+(2)-6\neq 0$

$P(-2)=2{ (-2) }^{ 2 }+(-2)-6=0$

$(x-1).(x+1).(x+2).(2x-3)$

$2x-3=2(x-\frac { 3 }{ 2 } )$

$2{ x }^{ 4 }+{ x }^{ 3 }-8{ x }^{ 2 }-x+6=2(x-1).(x+1).(x+2).(x-\frac { 3 }{ 2 } )$

$x=1,\quad x=-1,\quad x=-2,\quad x=\frac { 3 }{ 2 }$

Solution of exercise 10

$2{ x }^{ 3 }-7{ x }^{ 2 }+8x-3$

$P(1)=2{ (1) }^{ 3 }-7{ (1) }^{ 2 }+8(1)-3=2-7+8-3=0$

$(x-1).(2{ x }^{ 2 }-5x+3)$

$P(1)=2{ (1) }^{ 2 }-5(1)+3=2-5+3=0$

$(x-1).(x-1).(2x-3)={ (x-1) }^{ 2 }.2(x-\frac { 3 }{ 2 } )=2{ (x-1) }^{ 2 }(x-\frac { 3 }{ 2 } )$

$x=\frac { 3 }{ 2 } ,\quad x=1$

Solution of exercise 11

${ x }^{ 3 }-{ x }^{ 2 }-4$

$\{ \quad \pm 1,\quad \pm 2,\quad \pm 4\quad \}$

$P(1)={ (1) }^{ 3 }-{ (1) }^{ 2 }-4=1-1-4\neq 0$

$P(-1)={ (-1) }^{ 3 }-{ (-1) }^{ 2 }-4=-1-1-4\neq 0$

$P(2)={ (2) }^{ 3 }-{ (2) }^{ 2 }-4=8-4-4=0$

$(x-2).({ x }^{ 2 }+x+2)$

${ x }^{ 2 }+x+2=0$

$a{ x }^{ 2 }+bx+c=0$

$a=1,\quad b=1,\quad c=2$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(1)+\sqrt { { (1 })^{ 2 }-4(1)(2) } }{ 2(1) } ,\frac { -(1)-\sqrt { { { (1 }) }^{ 2 }-4(1)(2) } }{ 2(1) }$

$x=\quad \frac { 1+\sqrt { { 1 }-8 } }{ 2 } ,\frac { 1-\sqrt { { 1 }-8 } }{ 2 }$

$x=\quad \frac { 1+\sqrt { -7 } }{ 2 } \notin R,\frac { 1-\sqrt { -7 } }{ 2 } \notin R$

$(x-2).({ x }^{ 2 }+x+2)$

$x = 2$

Solution of exercise 12

${ x }^{ 3 }+3{ x }^{ 2 }-4x-12$

$\{ \pm 1,\quad \pm 2,\quad \pm 3,\quad \pm 4,\quad \pm 6\quad ,\pm 12\}$

$P(1)={ (1) }^{ 3 }+3{ (1) }^{ 2 }-4(1)-12=1+3-4-12\neq 0$

$P(-1)={ (-1) }^{ 3 }+{ 3(-1) }^{ 2 }-4(-1)-12=-1+3+4-12\neq 0$

$P(2)={ (2) }^{ 3 }+3{ (2) }^{ 2 }-4(2)-12=8+12-8-12=0$

$(x-2).({ x }^{ 2 }+5x+6)$

${ x }^{ 2 }+5x+6=0$

$a{ x }^{ 2 }+bx+c=0$

$a=1,\quad b=5,\quad c=6$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(5)+\sqrt { { (5 })^{ 2 }-4(1)(6) } }{ 2(1) } ,\frac { -(5)-\sqrt { { { (5 }) }^{ 2 }-4(1)(6) } }{ 2(1) }$

$x=\quad \frac { -5+\sqrt { { 25 }-24 } }{ 2 } ,\frac { -5-\sqrt { { 25 }-24 } }{ 2 }$

$x=\quad \frac { -5+\sqrt { 1 } }{ 2 } ,\frac { -5-\sqrt { 1 } }{ 2 }$

$x=\quad \frac { -5+1 }{ 2 } ,\frac { -5-1 }{ 2 }$

$x=\quad \frac { -4 }{ 2 } ,\frac { -6 }{ 2 }$

$x=-2,\quad x=-3$

$(x-2).(x+2).(x+3)$

$x=2, \quad x=-2, \quad x=-3$

Solution of exercise 13

${ 6x }^{ 3 }+7{ x }^{ 2 }-9x+2$

$\{ \pm 1,\quad \pm 2,\}$

$P(1)={ 6(1) }^{ 3 }+7{ (1) }^{ 2 }-9(1)+2=6+7-9+2\neq 0$

$P(-1)={ 6(-1) }^{ 3 }+7{ (-1) }^{ 2 }-9(-1)+2=-6+7+9+2\neq 0$

$P(2)={ 6(2) }^{ 3 }+7{ (2) }^{ 2 }-9(2)+2=48+28-18+2\neq 0$

$P(-2)={ 6(-2) }^{ 3 }+7{ (-2) }^{ 2 }-9(-2)+2=-48+28-18+2=0$

$(x+2).({ 6x }^{ 2 }-5x+1)$

${ 6x }^{ 2 }-5x+1=0$

$a{ x }^{ 2 }+bx+c=0$

$a=6,\quad b=-5,\quad c=1$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(-5)+\sqrt { { (5 })^{ 2 }-4(6)(1) } }{ 2(6) } ,\frac { -(-5)-\sqrt { { { (5 }) }^{ 2 }-4(6)(1) } }{ 2(6) }$

$x=\quad \frac { 5+\sqrt { { 25 }-24 } }{ 12 } ,\frac { 5-\sqrt { { 25 }-24 } }{ 12 }$

$x=\quad \frac { 5+\sqrt { 1 } }{ 12 } ,\frac { 5-\sqrt { 1 } }{ 12 }$

$x=\quad \frac { 5+1 }{ 12 } ,\frac { 5-1 }{ 12 }$

$x=\quad \frac { 6 }{ 12 } ,\frac { 4 }{ 12 }$

$x=\frac { 1 }{ 2 } \quad x=\frac { 1 }{ 3 }$

$6(x+2).(x-\frac { 1 }{ 2 } ).(x-\frac { 1 }{ 3 } )$

$x=-2,\quad x=\frac { 1 }{ 2 } ,\quad x=\frac { 1 }{ 3 }$

Solution of exercise 14

$9{ x }^{ 4 }-4{ x }^{ 2 }$

${ x }^{ 2 }(9{ x }^{ 2 }-4)$

${ x }^{ 2 }(3x+2)(3x-2)$

${ x }^{ 2 }[{ (3x) }^{ 2 }-{ (2) }^{ 2 }]$

${ x }^{ 2 }(9{ x }^{ 2 }-4)$

$9{ x }^{ 4 }-4{ x }^{ 2 }$

$x^{ 5 }+20{ x }^{ 3 }+100x$

$x(x^{ 4 }+20x^{ 2 }+100)$

$x{ (x^{ 2 }+10) }^{ 2 }$

$3x^{ 5 }-18{ x }^{ 3 }+27x$

$3x(x^{ 4 }-6x^{ 2 }+9)$

$3x{ ({ x }^{ 2 }-3) }^{ 2 }$

$2{ x }^{ 3 }-50x$

$2x({ x }^{ 2 }-25)$

$2x({ x }^{ 2 }-{ 5 }^{ 2 })$

$2x[(x-5)(x+5)]$

$2{ x }^{ 5 }-32x$

$2x({ x }^{ 4 }-16)$

$2x[{ ({ x }^{ 2 }) }^{ 2 }-{ 4 }^{ 2 }]$

$2x[({ x }^{ 2 }-4)({ x }^{ 2 }+4)]$

$2x[({ x }^{ 2 }-{ 2 }^{ 2 })({ x }^{ 2 }+4)]$

$2x[({ x }-{ 2 })({ x }+{ 2 })({ x }^{ 2 }+4)]$

$2{ x }^{ 2 }+x-28=0$

$a{ x }^{ 2 }+bx+c=0$

$a=2,\quad b=1,\quad c=-28$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(1)+\sqrt { { (1 })^{ 2 }-4(2)(-28) } }{ 2(2) } ,\frac { -(1)-\sqrt { { { (1 }) }^{ 2 }-4(2)(-28) } }{ 2(2) }$

$x=\quad \frac { -1+\sqrt { { 1 }+224 } }{ 4 } ,\frac { -1-\sqrt { { 1 }+224 } }{ 4 }$

$x=\quad \frac { -1+\sqrt { 225 } }{ 4 } ,\frac { -1-\sqrt { 225 } }{ 4 }$

$x=\quad \frac { -1+15 }{ 4 } ,\frac { -1-15 }{ 4 }$

$x=\quad \frac { 14 }{ 4 } ,\frac { -16 }{ 4 }$

$x=\frac { 7 }{ 2 } \quad x=-4$

$2{ x }^{ 2 }+x-28=2(x+4)(x-\frac { 7 }{ 2 } )$

Solution of exercise 15

$\frac { 2 }{ 5 } { x }^{ 5 }-\frac { 6 }{ 5 } { x }^{ 4 }+\frac { 14 }{ 15 } { x }^{ 2 }$

$\frac { 2 }{ 5 } { x }^{ 2 }({ x }^{ 3 }-3{ x }^{ 2 }+\frac { 7 }{ 3 } )$

$2xy-4x-3y+6$

$2x(y-2)-3(y-2)$

$(2x-3)(y-2)$

$25{ x }^{ 2 }-1$

${ (5x) }^{ 2 }-{ 1 }^{ 2 }$

$(5x-1)(5x+1)$

$36{ x }^{ 2 }-49$

${ (6x) }^{ 2 }-7^{ 2 }$

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

$5{ x }^{ 2 }-10x+5$

$5({ x }^{ 2 }-2x+1)$

$5[{ (x) }^{ 2 }-2(x)(1)+{ 1 }^{ 2 }]$

$5{ (x-1) }^{ 2 }$

${ x }^{ 2 }-6x+9$

${ x }^{ 2 }-2(x)(3)+{ 3 }^{ 2 }$

${ (x-3) }^{ 2 }$

${ x }^{ 2 }-20x+100$

${ x }^{ 2 }-2(x)(10)+{ 10 }^{ 2 }$

${ (x-10) }^{ 2 }$

${ x }^{ 2 }+10x+25$

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

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

${ x }^{ 2 }+14x+49$

${ x }^{ 2 }+2(x)(7)+{ 7 }^{ 2 }$

${ (x+7) }^{ 2 }$

${ x }^{ 3 }-4{ x }^{ 2 }+4x$

$x({ x }^{ 2 }-4x+4)$

$x({ x }^{ 2 }-2(x)(2)+{ 2 }^{ 2 })$

$x{ (x-2) }^{ 2 }$

$3{ x }^{ 7 }-27x$

$3x({ x }^{ 6 }-9)$

$3x[{ ({ x }^{ 3 }) }^{ 2 }-{ 3 }^{ 2 }]$

$3x{ ({ x }^{ 3 }-3) }({ x }^{ 3 }+3)$

${ x }^{ 2 }-11x+30=0$

$a{ x }^{ 2 }+bx+c=0$

$a=1,\quad b=-11,\quad c=30$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(-11)+\sqrt { { (-11 })^{ 2 }-4(1)(30) } }{ 2(1) } ,\frac { -(-11)-\sqrt { { { (-11 }) }^{ 2 }-4(2)(30) } }{ 2(1) }$

$x=\quad \frac { 11+\sqrt { { 121 }-120 } }{ 2 } ,\frac { 11-\sqrt { { 121 }-120 } }{ 2 }$

$x=\quad \frac { 11+\sqrt { 1 } }{ 2 } ,\frac { 11-\sqrt { 1 } }{ 2 }$

$x=\quad \frac { 11+1 }{ 2 } ,\frac { 11-1 }{ 2 }$

$x=\quad \frac { 12 }{ 2 } ,\frac { 10 }{ 2 }$

$x=6\quad x=5$

${ x }^{ 2 }-11x+30=(x-5)(x-6)$

$3{ x }^{ 2 }-10x+3=0$

$a{ x }^{ 2 }+bx+c=0$

$a=3,\quad b=-10,\quad c=3$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(-10)+\sqrt { { (-10 })^{ 2 }-4(3)(3) } }{ 2(3) } ,\frac { -(-10)-\sqrt { { { (-10 }) }^{ 2 }-4(3)(3) } }{ 2(3) }$

$x=\quad \frac { 10+\sqrt { { 100 }-36 } }{ 6 } ,\frac { 10-\sqrt { { 100 }-36 } }{ 6 }$

$x=\quad \frac { 10+\sqrt { 64 } }{ 6 } ,\frac { 10-\sqrt { 64 } }{ 6 }$

$x=\quad \frac { 10+8 }{ 6 } ,\frac { 10-8 }{ 6 }$

$x=\quad \frac { 18 }{ 6 } ,\frac { 2 }{ 6 }$

$x=3\quad x=\frac { 1 }{ 3 }$

$3{ x }^{ 2 }-10x+3=(x-3)(x-\frac { 1 }{ 3 } )$

$2{ x }^{ 2 }-x-1=0$

$a{ x }^{ 2 }+bx+c=0$

$a=2,\quad b=-1,\quad c=-1$

$x=\quad \frac { -(-1)+\sqrt { { (-1 })^{ 2 }-4(2)(-1) } }{ 2(2) } ,\frac { -(-1)-\sqrt { { { (-1 }) }^{ 2 }-4(2)(-1) } }{ 2(2) }$

$x=\quad \frac { 1+\sqrt { { 1 }+8 } }{ 4 } ,\frac { 1-\sqrt { { 1 }+8 } }{ 4 }$

$x=\quad \frac { 1+\sqrt { 9 } }{ 4 } ,\frac { 1-\sqrt { 9 } }{ 4 }$

$x=\quad \frac { 1+3 }{ 4 } ,\frac { 1-3 }{ 4 }$

$x=\quad \frac { 4 }{ 4 } ,\frac { -2 }{ 4 }$

$x=1\quad x=-\frac { 1 }{ 2 }$

$2{ x }^{ 2 }-x-1=(x-1)(x+\frac { 1 }{ 2 } )$

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Radka MacKillop

May 2020

There are two mistakes in Exercise 5. The factors are right but the last line of the last two problems is wrong. If you multiply the right side, you don’t get the expression in the left side. The number coefficients (respectively 3 and 2) are missing.

Factoring Polynomials Worksheet

Factor and Calculate the Roots of the Following Polynomials

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Exercise 9

Exercise 10

Exercise 11

Exercise 12

Exercise 13

Exercise 14

Exercise 15

Solution of exercise 1

Solution of exercise 2

Solution of exercise 3

Solution of exercise 4

Solution of exercise 5

Solution of exercise 6

Solution of exercise 7

Solution of exercise 8

Solution of exercise 9

Solution of exercise 10

Solution of exercise 11

Solution of exercise 12

Solution of exercise 13

Solution of exercise 14

Solution of exercise 15

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