分析:運用因式分解將x4+7x3+8x2-13x+15轉(zhuǎn)化為x2(x2+2x)+5X3+8x2-13x+15,將x2+2x做為整體代入上式,這樣就降低了x的次數(shù),并進(jìn)一步轉(zhuǎn)化為5x(x2+2x)+x2-13x+15,再將x2+2x做為整體代入5x(x2+2x)+x2-13x+15式,此時原式轉(zhuǎn)化為x2+2x+15,又出現(xiàn)x2+2x,再代入,至此問題解決.
解答:解:x2+2x=3
(x-1)(x+3)=0
則x=1或-3
所以x4+7x3+8x2-13x+15=x2(x2+2x)+5x3+8x2-13x+15
=x2×3+5x3+8x2-13x+15
=5x3+11x2-13x+15
=5x(x2+2x)+x2-13x+15
=15x+x2-13x+15
=x2+2x+15
=3+15
=18
故答案為18
點評:本題考察因式分解.本題解決的關(guān)鍵是將x2+2x整體逐級代入x4+7x3+8x2-13x+15變化后的式子,降低了x的次數(shù),使問題最終得以解決.